The value of the frequency for grade C from given table of final grade is 8.
As given,
Given frequency table is:
Grade: A B C D F
Frequency: 4 7 ? 3 2
Relative Frequency: 16.70% 29.20% 33.30% 12.50% 8.30%
Cumulative Frequency: 4 11 19 22 24
Given records: A, A, A, A, B, B, B, B, B, B, B, C, C, C,C, C, C, C, C,D,D,D, F,F
Here : frequency of grade C is 8 times
or Cumulative frequency (19-11=8).
Therefore, the value of the frequency for grade C from the given table of final grade is 8.
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If Point B is is the midpoint of line AC how do I dove the correct way to find B?
The coordinates of point B, (x, y) is given by (x, y) = ((a₁ + c₁)/2, (a₂ + c₂)/2)
Midpoint of a lineFrom the question, we are to determine how to find the coordinates of B
From the given information,
Point B is the midpoint of line AC
If the coordinates of point A are (a₁, a₂) and the coordinates of point C are (c₁, c₂)
Then,
The coordinates of the midpoint (x, y) is given by
(x, y) = ((a₁ + c₁)/2, (a₂ + c₂)/2)
Hence, the coordinates of point B, (x, y) is (x, y) = ((a₁ + c₁)/2, (a₂ + c₂)/2)
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HELP WITH ALLLLL PLEASE THIS IS ALGEBRA II BTWWW:))
Answer:
[tex]\textsf{23.} \quad c)\;\;y=-2x+7[/tex]
[tex]\textsf{24.} \quad 12[/tex]
[tex]\textsf{25.} \quad -2.33\:\:\sf (2\:d.p.)[/tex]
Step-by-step explanation:
Question 23Given equation:
[tex]y=\dfrac{1}{2}x-1[/tex]
If two lines are perpendicular to each other, the slopes are negative reciprocals.
Therefore, the slope of a line perpendicular to the given equation is -2.
Substitute the found slope and the given point (3, 1) into the point-slope formula to find the equation of the line:
[tex]\implies y-y_1=m(x-x_1)[/tex]
[tex]\implies y-1=-2(x-3)[/tex]
[tex]\implies y-1=-2x+6[/tex]
[tex]\implies y=-2x+7[/tex]
Question 24Define the variables:
Let x = number of $20 bills.Let y = number of $50 bills.Given information:
Total amount cashed = $390Total number of bills = 15Create two equations with the given information:
[tex]\begin{cases}x+y=15\\20x+50y=390\end{cases}[/tex]
Solve the first equation for y:
[tex]\implies y=15-x[/tex]
Substitute the found expression for y into the second equation and solve for x:
[tex]\implies 20x+50(15-x)=390[/tex]
[tex]\implies 20x+750-50x=390[/tex]
[tex]\implies -30x+750=390[/tex]
[tex]\implies -30x=-360[/tex]
[tex]\implies x=12[/tex]
Therefore, Kerry received 12 twenty-dollar bills.
Question 25Given expression:
[tex]\dfrac{6^2-4^2}{-10+\sqrt{2}}[/tex]
Following the order of operations (PEMDAS), simplify the numerator:
[tex]\implies \dfrac{36-16}{-10+\sqrt{2}}[/tex]
[tex]\implies \dfrac{20}{-10+\sqrt{2}}[/tex]
Calculate the square root:
[tex]\implies \dfrac{20}{-10+1.414213...}[/tex]
Simplify the denominator:
[tex]\implies \dfrac{20}{-8.5857864...}[/tex]
Divide the numerator by the denominator:
[tex]\implies -2.3294313...[/tex]
Therefore:
[tex]\implies \dfrac{6^2-4^2}{-10+\sqrt{2}}=-2.33\:\: \sf (2\:d.p.)[/tex]
[tex]-8(x-4)-3\leq 5[/tex]
In (triangle) ABC, AB is extended to D. If m/ACB = 8x, m/CAB = 5x+10, and, m/CBD=7x+34, what is the value of x?
Step-by-step explanation:
the sum of fmeasure angle of triangle is 180°
so <A + < B + <C = 180
6x - 24 = 0
x = 4
Given the function 7x+2y=6, rearrange the equation so that x is the independent variable.
Two expressions with an equal sign is equation, The given function is 7x+2y=6, The equation when x is independent variable is x=(6-2y)/7.
What is equation?Two expressions with an equal sign in between is called as equation.
The given equation is 7x+2y=6.
In the given equation x and y are two variables.
According to the question, to make x as independent variable we have to perform few operation in equation.
7x+2y=6.
Subtract 2y on both sides
7x+2y-2y=6-2y
7x=6-2y
Divide both sides by 7
x=(6-2y)/7.
Therefore the independent variable x is (6-2y)/7.
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x+1=7x-5 need to find x
The value of "x" which will satisfy the equation [(x + 1) = (7x - 5)] is 1.
As per the question statement, we are provided with a linear equation
[(x + 1) = (7x - 5)].
We are required to determine the value of "x" for which, the above mentioned equation will be satisfied.
To solve this question, we simply need to rearrange our given equation, and bring the variable terms to one side, and the integer terms to another, and then solve for "x", as shown below.
{[(x+1)=(7 x-5)] }
or,(5+1)=(7 x-x)
or,(7-1) x=(5-1)
or, 6 x=6
or,x=6/5
or,x=1
That is, the value of "x" which will satisfy the equation [(x + 1) = (7x - 5)] is 1.
Equation: In mathematics, an equation is a formula or a statement, that expresses the relation of equality between two expressions, by connecting them with the "equal to" sign.To learn more about Equations, click on the link below
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Ali bought three pieces of fruit with the following weights, 8.712 grams, 14.125 grams, and 21.926 grams. What is the best estimate for the total weight of the fruit?
The total estimated weight of the three pieces of the fruit is 45 g.
What is addition?Combining things and counting them as one big group is done through addition. Addition is the process of combining two or more integers in mathematics. The addends are the numbers that are added, whereas the total is the result of the operation.
Given: The weight of three pieces"
8.712 g
14.125 g
21.926 g
The estimated weight of the three fruits will be:
8.712~ 9 g
14.125 ~ 14 g
21.926 ~ 22 g
So, the total estimated weight of the three pieces of the fruit = 9 + 14 + 22
The total estimated weight of the three pieces of the fruit = 45 g
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How are evidence and counterexamples used in proofs?
In a direct proof, evidence is used to
Reset
On the other hand, a counterexample is a single example that
k
Next
The required evidence is used to give stand to proofs while counterexamples are used to prove the proof false.
Given that,
To specify how are evidence and counterexamples used in proofs.
In mathematics, it deals with numbers of operations according to the statements. There are four major arithmetic operators, addition, subtraction, multiplication and division,
Here,
proving
a² = 1
Let assume a = 1
When squaring both sides
a² = 1
Conture an example of the proof a² = 1 to prove a = 1
a² = 1
a = ± 1
a ≠ 1 alone
Thus, the required evidence is used to give stand to proofs while counterexamples are used to prove the proof false.
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500 g of a radioactive element is decaying exponentially. After 8 days 358 g of the element is left.
a. Write a function in the form y = yo ekt giving the number of grams of the element after t days.
358 f(t)
b. Write the function from part a in the form y=Yo
500
c. Use the answer from part a to find the half-life of the element.
a. The exponential equation in the form y=yoekt is
(Round to three decimal places as needed.)
From the situation described, we have that:
a) The exponential function is: [tex]y(t) = 500e^{-0.041759389t}[/tex]
b) The rule is: [tex]y(t) = y(0)\left(\frac{358}{500}\right)^{\frac{t}{8}}[/tex]
c) The half-life of the element is of 16.599 days.
What is the exponential function?The exponential function for the substance's amount after t days is given by:
[tex]y(t) = y(0)e^{-kt}[/tex]
For this problem, we have that:
y(0) = 500, y(8) = 358.
Hence we can solve for k as follows:
[tex]y(t) = y(0)e^{-kt}[/tex]
[tex]358 = 500e^{-8k}[/tex]
[tex]e^{-8k} = \frac{358}{500}[/tex]
[tex]\ln{e^{-8k}} = \ln{0.716}[/tex]
-8k = ln(0.716)
k = -ln(0.716)/8
k = 0.041759389.
Hence:
[tex]y(t) = 500e^{-0.041759389t}[/tex]
For item b, 358/500 of the substance is the amount remaining each period of 8 days, hence f(t) = t/8 and the rule is given by:
[tex]y(t) = y(0)\left(\frac{358}{500}\right)^{\frac{t}{8}}[/tex]
For item c, using the rule from item a, we have to find t for which y(t) = 0.5y(0), hence:
[tex]y(t) = y(0)e^{-0.041759389t}[/tex]
[tex]0.5y(0) = y(0)e^{-0.041759389t}[/tex]
[tex]e^{-0.041759389t} = 0.5[/tex]
[tex]\ln{e^{-0.041759389t}} = \ln{0.5}[/tex]
-0.041759389t = ln(0.5)
t = -ln(0.5)/0.041759389
t = 16.599.
The half-life of the element is of 16.599 days.
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21/09/22
un articulo regulamente cuesta $85 pesos esta a
la venta con un descuento del 30% del precio regular
¿Cual es el precio de venta?
Can you help me with this problem.
According to the given figure, the length of interval AB can be denoted by AB =6.40312.
Let's consider the intersecting point of A and b
is p
And
AB is c
AP is 6-2= 4 units = a
BP is 6-1 = 5 units = b
Applying the Pythagorean theorem :
Pythagorean theorem,
the well-known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)—or, in familiar algebraic notation, a2 + b2 = c2
4^2 +5^2 = 16+25
=41
c^2 = 41
C= root of 41 = 6.40312
AB is 6.40312
Algebraic proofs
The theorem can be proved algebraically using four copies of the same triangle arranged symmetrically around a square with side c, as shown in the lower part of the diagram.[5] This results in a larger square, with side a + b and area (a + b)2. The four triangles and the square side c must have the same area as the larger square,
{\displaystyle (b+a)^{2}=c^{2}+4{\frac {ab}{2}}=c^{2}+2ab,}{\displaystyle (b+a)^{2}=c^{2}+4{\frac {ab}{2}}=c^{2}+2ab,}
giving
{\displaystyle c^{2}=(b+a)^{2}-2ab=b^{2}+2ab+a^{2}-2ab=a^{2}+b^{2}.}{\displaystyle c^{2}=(b+a)^{2}-2ab=b^{2}+2ab+a^{2}-2ab=a^{2}+b^{2}.}
A similar proof uses four copies of a right triangle with sides a, b and c, arranged inside a square with side c as in the top half of the diagram.[6] The triangles are similar with area {\displaystyle {\tfrac {1}{2}}ab}{\tfrac {1}{2}}ab, while the small square has side b − a and area (b − a)2. The area of the large square is therefore
{\displaystyle (b-a)^{2}+4{\frac {ab}{2}}=(b-a)^{2}+2ab=b^{2}-2ab+a^{2}+2ab=a^{2}+b^{2}.}{\displaystyle (b-a)^{2}+4{\frac {ab}{2}}=(b-a)^{2}+2ab=b^{2}-2ab+a^{2}+2ab=a^{2}+b^{2}.}
But this is a square with side c and area c2, so
c^{2}=a^{2}+b^{2}
c^{2}=a^{2}+b^{2}
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The length of the interval AB can be expressed as AB =6.40312 using the accompanying figure.
Consider the location where A and b intersect.
is p
And
AB is c
AP is 6-2/4 units, or an.
BP equals 6-1 = 5 units + b.
Making use of the Pythagorean theorem
Pythagorean principle,
the well-known geometric theorem that states that the hypotenuse of a right triangle—the side across from the right angle—is equal to the sum of its legs' squares, or, in familiar algebraic form, a2 + b2 = c2.
4^2 +5^2 = 16+25
=41
c^2 = 41
41 divided by C equals 6.40312
AB is 6.40312
mathematic proofs
As indicated in the lower portion of the diagram, the theorem can be proven algebraically by placing four identical triangles symmetrically around a square with side c. [5] Consequently, a larger square is formed, with side a + b and area (a + b)2. The area of the larger square must be shared by the four triangles and the square side c.
{\displaystyle (b+a)^{2}=c^{2}+4{\frac {ab}{2}}=c^{2}+2ab,}{\displaystyle (b+a)^{2}=c^{2}+4{\frac {ab}{2}}=c^{2}+2ab,}
giving
{\displaystyle c^{2}=(b+a)^{2}-2ab=b^{2}+2ab+a^{2}-2ab=a^{2}+b^{2}.}{\displaystyle c^{2}=(b+a)^{2}-2ab=b^{2}+2ab+a^{2}-2ab=a^{2}+b^{2}.}
A similar proof uses four copies of a right triangle with sides a, b and c, arranged inside a square with side c as in the top half of the diagram.[6] The triangles are similar with area {\displaystyle {\tfrac {1}{2}}ab}{\tfrac {1}{2}}ab, while the small square has side b − a and area (b − a)2. The area of the large square is therefore
{\displaystyle (b-a)^{2}+4{\frac {ab}{2}}=(b-a)^{2}+2ab=b^{2}-2ab+a^{2}+2ab=a^{2}+b^{2}.}{\displaystyle (b-a)^{2}+4{\frac {ab}{2}}=(b-a)^{2}+2ab=b^{2}-2ab+a^{2}+2ab=a^{2}+b^{2}.}
But this is a square with side c and area c2, so
c^{2}=a^{2}+b^{2}
c^{2}=a^{2}+b^{2}
A bag with 8 marbles is shown below. 2 marbles are red and 6 are blue.
The probability that it is blue or red is 1
How to determine the probability?The given parameters are
Marbles = 8
Red = 2
Blue = 6
The probability that the selected marble is blue or red is calculated as
P = P(Blue) + P(Red)
This gives
P = 6/8 + 2/8
Evaluate the sum
P = 1
Hence, the probability that it is blue or red is 1
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Complete question
A bag with 8 marbles is shown below. 2 marbles are red, and 6 are blue. A marble is chosen from the bag at random. What is the probability that it is blue or red?
Write your answer as a fraction or a whole number.
i need help figuring this out.
Answer:
b
Step-by-step explanation:
Please help I need it for school
Answer:
this game needs step to step explanation
There are three bowling balls. The balls average 14 pounds. The first ball weighs 12 pounds. The second ball weighs 15 pounds. How much does the third ball weigh?
Enter your answer in the box as a whole number.
The third ball weighs 15 pounds.
Given that there are three bowling balls, the balls average 14 pounds.
The first ball weighs 12 pounds.
The second ball weighs 15 pounds.
We need to find the weight of the third ball.
To find the weight of the third ball, you can use the average weight formula:
Average weight = (Sum of weights) / (Number of balls)
Given that the average weight of the three balls is 14 pounds and you know the weights of the first two balls (12 pounds and 15 pounds), you can find the weight of the third ball (let's call it "x") using the formula:
14 = (12 + 15 + x) / 3
Now, solve for x:
14 = (27 + x) / 3
Multiply both sides by 3 to eliminate the denominator:
42 = 27 + x
Subtract 27 from both sides:
x = 42 - 27
x = 15
Therefore, the weight of the third ball is 15 pounds.
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Write a piecewise definition for the function and sketch its graph.
The piecewise definition for the function is
f(x) = 3x^3, x < 0f(x) = -3x^3; x >= 0How to write a piecewise definition for the function and sketch its graph?The definition of the function is given as:
f(x) = |3x^3|
Remove the absolute bracket in the above function
So, we have:
f(x) = 3x^3 and f(x) = -3x^3
The domain of the above functions are
x < 0 and x >=0
Next, we plot the graph of the function
See attachment for the graph of the piecewise definition for the function
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rewrite each expression by factoring out the coefficient of variable -2x+3
By factoring out the coefficient of variable, -2x+3 can be written as [tex]-2(x-\frac{3}{2})[/tex].
Factoring out means isolating a common factor from an expression. A number or a variable that is multiplied by another variable in the expression forms the coefficient. The factoring out of the coefficient of variable means taking the factor of the coefficient part of the variable term.
Given data is an expression -2x+3.
-2 is the coefficient of variable x. So, factoring out the (-2) from -2x+3:
It can be written as [tex]-2(x-\frac{3}{2} )[/tex]
Hence, -2x+3 can be written [tex]-2(x-\frac{3}{2})[/tex] by factoring out the coefficient of variable.
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100 points
f(x+h) - f(x) = -3hx² + 7hx + 4h²x - 5h² - 3h³
find f'(x)
The required differentiation of f(x) is f'(x) = 6hx -7h -4h² - f'(x + h)
Given that,
To determine the differentiation f'(x)
f(x+h) - f(x) = -3hx² + 7hx + 4h²x - 5h² - 3h³ is given,
Differentiation is defined as the instantaneous rate of change of a particular quantity with respect to another.
Here,
f(x + h) - f(x) = -3hx² + 7hx + 4h²x - 5h² - 3h³
differentiate with respect to x
d[f(x + h)]/dx - df(x)/ dx = -3hdx² /dx + 7hdx /dx + 4h²dx/dx - d(5h² + 3h³)/dx
f'(x + h) - f'(x) = -6hx + 7h + 4h²
f'(x) = 6hx -7h -4h² - f'(x + h)
Thus, required differentiation of f(x) is f'(x) = 6hx -7h -4h² - f'(x + h).
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find the width and height of an newer 90 inch televison whose aspect ratio is 16:9
The width and height of a newer 90 inch televison whose aspect ratio is 16:9 are 57.6 and 32.4 inches
How to find the width and height of an newer 90 inch televison whose aspect ratio is 16:9?The given parameters are
Aspect ratio = 16 : 9
Size = 90 inches
The width is calculated as
Width = 16/(16 + 9) * 90
Evaluate
Width = 57.6
This means that
height = 9/(16 + 9) * 90
Evaluate
height = 32.4
Hence, the width and height of a newer 90 inch televison whose aspect ratio is 16:9 are 57.6 and 32.4 inches
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Need some help with this
Using the Pythagorean theorem what is A? A=? b=8 c=10
Answer:
A = 6
Step-by-step explanation:
Look at the picture
A particle moves along a number line according
to the following instructions.
Start at position 5
Move left 4 units
Move right 8 units
Move right 6 units
Move left 13 units
How far is the particle from the initial starting point?
A. 2
B. 3
C. 8
D. 26
E. 36
-
-
-
-
-
at the end the particle is at 3 units from the starting point.
How far is the particle from the initial starting point?
Here we have one-dimensional motion, so the position of the particle is given by scalars.
We know that the particle starts at position 5. If it moves to the left, we subtract the value, if it moves to the right, we add the value.
Here we have the sequence of motions:
Move left 4 unitsMove right 8 unitsMove right 6 unitsMove left 13 unitsThen the final position will be at:
5 - 4 + 8 + 6 - 13 = 2
Then the final position is at 2, and the initial position is at 5, the difference between these is: 5 - 2 = 3
Then, at the end the particle is at 3 units from the starting point.
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Slove this question!!!!!!!!!!!!!!
Answer:
247/8
Step-by-step explanation:
Length of gestation < 20 weeks 20 − 27 weeks 28 − 36 weeks > 36 weeks Probability .0004 .0059 .0855 .9082 Furthermore, the conditional probability of low birth weight, given that length of gestation is < 20 weeks is .540 while this probability is .813, given that length of gestation is 20−27 weeks, is .379 when given that length of gestation is 28 − 36 weeks, and this probability is .035, given that length of gestation is > 36 weeks. (a) What is the probability that an infant has a low birth weight? (b) Show that the events {length of gestation ≤ 27 weeks} and {low birth weight} are not independent. (c) What is the probability of having a length of gestation ≤ 36 weeks given that a child is low birth weight?
(a) The probability of an infant has a low birth weight is equal to 1540.
(b) Won't find a probability that the event is not less than 27 and are not independent. So we need to answer that we will have to find a low both by given less than equal to the 27 point.
(c) Probability in less than equal to 36 and 8, given problem the 1 now for the probability the we have found already equal to 0692 point and for the time it will be the sum of the 3 of them.
The length of the gestation 123 under 4, and the first 1 will be small and then the came 3 weeks, and this 1 will be from the 22. The 27 weeks- and this will be the 28 to the 36 weeks- this will be the greater than the 36 weeks and we have the probability for each of them, for the first 1 will be. The 1 will be 59 point. This 1 will be the 855, and this 1 will be the ebon 9 to 82 and the next 1. We will have the call this. The means that the will be the frame is not the lowest weight, so condition will be low, and this 1 will be no more, which makes it easier now and no more no more. It will be, and this will be no more.
It will again now and then no more so from the question here even to win be less than 20, with the probability low weight equal to 1540, the common among 1 minus that equal to 46 point the next 1 for the 2227 point, the probability equal to the 1, is 13, the common 1 equals to 1 minus 0 upon , 131 minus upon 13 ease and the next 1 will on 3791 minus 379 to 621, and the last 1 will be the 1351 minus 35 o 965, and that will be the summary of The question here the question: find the probability that an infant has a child, so we have the lobed and we want to root the. By going to the first plan, and this 1 . I got you this 1 and then got you. got the last 1 and then got to the low, so we multiply them like a pair and then we add them up. So if we do it, we should get the tiptoe. We keep going like this until the last value will be on 982 time 35, and if we compute it, we will have the 4 times 4 plus 59 times upon a 13 plus 3855 times 379 plus upon 9082 times 35 equal to 0692 point for the question B, won't you find a probability that the event is not less than 27 and are not independent. So we need to answer that we will have to find a low both by given less than equal to the 27 point, so it will be between the both of them here. So we will end up the first and second 1, so we will have. It will be equal to. I will try to find the probability, the al and the less than equal to 27 point. So we'll add up this 1 and this 1 together. So we get on 5 , 59 and 10 times 13, and if we compare the 54 plus 5910 on a 135127 point and then we want to find a probability of the time with the probability point probably the end. We have equal to the 1692 we found in the time when the problem is the last time question to 27 will be the sum of the all be the 59, and we can equal to the 59 times 4 equal to the 2152 times 10 to the power. Minus 6 and it's a different form, the 127 point, so the form they are not independent and for of question c 1251 is the probability of the testation will be less than equal to the thirty. Sixth, given that this 1 by the formula equal probability in less than equal to 36 and 8, given problem the 1 now for the probability the we have found already equal to 0692 point and for the time it will be the sum of the 3 of them here on together, so it will be easy adjacent to her .0692 l minus the last 1 here will be minus 982 times 35 and if we compute it, we go have on in 82 times 35006. So let me compute again. 0.9082100. .035 point. the son from here, so let me compare again time, 5 or so 59 times upon it: 13 plus 55 times 3 on 982 times upon 35, then this 1 equal to the point 6 p. This will be correct here, 692 point. So this 1 minus 82 times 5692 point and if we compute we can equal to the 0.54071.
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Which expression is equivalent to sum of quantity negative three and one third times n plus one sixth end quantity plus quantity one and three sixths times n minus three twelfths end quantity?
A: negative twenty nine sixths times n minus one twelfth
B: negative twenty nine sixths times n plus one twelfth
C: negative eleven sixths times n minus one twelfth
D: negative eleven sixths times n plus five twelfths
The equivalent expression is negative eleven sixths times n minus one twelfth. Option C
What are algebraic expressions?Algebraic expressions are expressions that are made up of terms, variables, factors, coefficients, constants.
They are also made up of mathematical operations.
Equivalent expressions are defined as expressions have the exact same solution but may have different arrangement of values.
Given the expression;
[ -3 1/ 3n + 1/ 6 } + { 1 3/ 6n - 3/ 12 }
simply the expression;
- 10/ 3n + 1/ 6 + 9/6n - 3/ 12
collect like terms
-10/ 3n + 9/ 6n + 1/ 6 - 3/ 12
-20n + 9n/6 + 2 - 3 /12
Add like terms
-11n/ 6 - 1/ 12
Thus, the equivalent expression is negative eleven sixths times n minus one twelfth. Option C
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Simplify (4.3 × 10−2)(5.12 × 10−10). Write the final answer in scientific notation.
2.2016 × 10^−11
2.2016 × 10^20
22.016 × 10^−12
2.2016 × 10^20
The scientific notation of (4.3 × 10⁻²)(5.12 × 10⁻¹⁰) on simplifying is 2.2016 × 10⁻¹¹ which is option (A).
Scientific notation could be a thanks to write terribly massive and really little numbers. There square measure 2 elements to scientific notation, a constant between one however but ten and also the power of 10. The constant features a whole half (left of the decimal point) and also the decimal half (right of the decimal point) referred to as the fixed-point part.It is given that (4.3 × 10⁻²)(5.12 × 10⁻¹⁰) .
On multiplying 4.3 by 5.12 , we get
4.3 × 5.12 = 22.016
Using exponent law for 10⁻² and 10⁻¹⁰ , we get
10⁻² × 10⁻¹⁰ = 10⁻²⁻¹⁰
= 10⁻¹²
Hence as a whole it is written as 22.016 × 10⁻¹².
In scientific notation , it is written as 2.2016 × 10⁻¹¹ .
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Gabrielle buys a plant that is 3 inches tall. It grows at a rate of 3 1 inches per month. The graph below represents the relationship between the number of months and the height of the plant.
The expression that represents the relationship between the number of months and the height of the plant will be 3 + 3.1m
How to illustrate the information?It should be noted that from the information, Gabrielle buys a plant that is 3 inches tall and the plant grows at a rate of 3.1 inches per month.
Therefore, the expression to show the relationship will be:
= 3 + (3.1 × m)
= 3 + 3.1m
Therefore, the expression that represents the relationship between the number of months and the height of the plant will be 3 + 3.1m.
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Answer:
2
Step-by-step explanation:
2/19=2:19
let f(x)=3x^2+7, g(x)=x^3+5x^2-3 and h(x)=7x^2+x a. what is the formula for t if t(x)=f(x)-g(x).
Solving the Question
We're given:
[tex]f(x)=3x^2+7[/tex][tex]g(x)=x^3+5x^2-3[/tex][tex]h(x)=7x^2+x[/tex][tex]t(x)=f(x)-g(x)[/tex]To solve for t(x), we simply plug in the expressions given for f(x) and g(x):
[tex]t(x)=f(x)-g(x)\\t(x)=(3x^2+7)-(x^3+5x^2-3)[/tex]
[tex]t(x)=3x^2+7-x^3-5x^2+3\\t(x)=-x^3+3x^2-5x^2+7+3\\t(x)=-x^3-2x^2+10[/tex]
Answer[tex]t(x)=-x^3-2x^2+10[/tex]
Simplify the expression.
-6(2 + x) =