How much volume (in cm3 ) is gained by a person who gains 11.0 lblb of pure fat?

Answer:

19.4

Step-by-step explanation:

The distance between the x coordinates of -4 and 15 would be 19 and the distance between the y coordinates of 2 and 6 would be 4.  We can use these to draw a right triangle.  The legs would be 19 and 4.  We would use the Pythagorean Theorem to find the hypotenuse.

[tex]a^{2}[/tex] + [tex]b^{2}[/tex] = [tex]c^{2}[/tex]

[tex]19^{2}[/tex] + [tex]4^{2}[/tex] = [tex]c^{2}[/tex]

361 + 16 = [tex]c^{2}[/tex]

377 =[tex]c^{2}[/tex]

[tex]\sqrt{377}[/tex] = [tex]\sqrt{c^{2} }[/tex]

19.4 ≈ c  That is our answer rounded to the nearest tenth

The coordinates of the midpoint of a segment with the given endpoints will be (-19, 5).

Let AB be the line segment and C be the mid-point of line segment AB. Let the coordinate of point A (x₁, y₁), the coordinate of point B (x₂, y₂), and the coordinate of the mid-point (x, y).

Then the coordinate of the mid-point of the line segment is given as,

(x, y) = [(x₁ + x₂) / 2, (y₁ + y₂) / 2]

The endpoints are A(-28, 8) and C(-10, 2).

Then the coordinates of the midpoint of a segment with the given endpoints will be

(x, y) = [(-28 - 10) / 2, (8 + 2) / 2]

(x, y) = (-38/2, 10/2)

(x, y) = (-19, 5)

The coordinates of the midpoint of a segment with the given endpoints will be (-19, 5).

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By substitution, the solution of the system of equations, x + 3y = 7 and 2x - 4y = 24, is (10 , -1).

A system of linear equations is a set of two or more equations which includes common variables. To solve system of equations, we must find the value of the unknown variables used in the equations that must satisfy both equations.

There are three methods that can be used to solve system of linear equations.

1. Elimination

2. Substitution

3. Graphing

Using substitution method, given two linear equations in x and y,

x + 3y = 7     (equation 1)

2x - 4y = 24     (equation 2)

Using the first equation, find the value of one variable, lets say x, in terms of the other.

x + 3y = 7     (equation 1)

x = 7 - 3y

Substitute the equation of x to the second equation.

2x - 4y = 24     (equation 2)

2(7 - 3y) - 4y = 24

14 - 6y - 4y = 24

Combining all terms containing the variable y on one side and the constants on the other side of the equality, and solving for y.

14 - 6y - 4y = 24

-6y - 4y = 24 - 14

-10y = 10

y = -1

Substitute the value of y in the equation of x.

x = 7 - 3y

x = 7 - 3(-1)

x = 7 - (-3)

x = 10

Hence, the solution of the given system of equations is (10 , -1).

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Answer:

D, since 7 is less than 10.

Step-by-step explanation:

Given,

X = 3/5, Y = 1/3, Z = 2×4/5

the given question is to find out the value of Z+X×Y so the value is stated as

14/5+3/5×1/3

Here, 14/5 will be remains same and you should multiply 3/5×1/3.

Let 14/5 be equation number (1)

The L.C.M for 3/5×1/3 is 15. then,

3/5×15+1/3×15

___________

15

3×3+1×5

= -------------

15

9+5/15 = 14/15.

Let 14/15 be equation number (2)

Combine equation number (1) and (2) then

14/5+14/15

L.C.M.of 14/5 and 14/15 is 15.

14/5×15+14/15×15

-------------------------

15

[tex] = \frac{42 + 14}{15} [/tex]

[tex] \frac{56}{15} [/tex]

The sum of the 9 terms of the given series -5-25-45- . . . . is -765

The given series is:

-5-25-45- . . .

and there are 9 terms.

Notice that the given series is an arithmetic series with:

first term, a(1) = -5

common difference, d = -25 - (-5) = -20

Recall the formula for the nth term of an arithmetic series:

a(n) = a(1) + (n - 1) . d

Substitute a(1) = -5, d = -20, and n = 9 into the formula:

a(9) = -5 + (9 - 1) . (-20)

a(9) = - 5 + (-160)

a(9) = -165

To find the sum of the series, use the sum formula:

S(n) = n/2 [(a(1) + a(n)]

with n = 9.

Hence,

S(9) = 9/2 [-5 + (-165)]

S(9) = -765

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The curl of the vector field is:

[tex]=z e^x \hat{\imath}-\left(y z e^x-7 x y e^z\right) \hat{\jmath}-7 x e^z \hat{k}[/tex]

So,

To find the curve:

[Vector Field] Set up [curl F]:

[tex]Curl F=\left|\begin{array}{ccc}\hat{1} & \hat{\jmath} & \hat{k} \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\7 x y e^z & 0 & y z e^x\end{array}\right|[/tex]

Simplify [3x3 Matrix Determinant] with [curl F]:

[tex]Curl F=\left|\begin{array}{cc}\frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\0 & y z e^x\end{array}\right| \hat{i}-\left|\begin{array}{cc}\frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\7 x y e^z & y z e^x\end{array}\right| \hat{j}+\left|\begin{array}{cc}\frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\7 x y e^z & 0\end{array}\right| \hat{k}[/tex]

Simplify [2x2 Matrix Determinant] with [curl F]:

[tex]Curl F=\left[\frac{\partial}{\partial y} y z e^x-\frac{\partial}{\partial z} 0\right] \hat{\mathrm{i}}-\left[\frac{\partial}{\partial x} y z e^x-\frac{\partial}{\partial z} 7 x y e^z\right] \hat{\mathrm{\jmath}}+\left[\frac{\partial}{\partial x} 0-\frac{\partial}{\partial y} 7 x y e^z\right] \hat{\mathrm{k}}[/tex]

The partial derivatives can be differentiated using the basic differentiation techniques listed above under "Calculus":

[tex]\begin{aligned}&\frac{\partial}{\partial y} y z e^x=z e^x \\&\frac{\partial}{\partial z} 0=0 \\&\frac{\partial}{\partial x} y z e^x=y z e^x \\&\frac{\partial}{\partial z} 7 x y e^z=7 x y e^z \\&\frac{\partial}{\partial x} 0=0 \\&\frac{\partial}{\partial y} 7 x y e^z=7 x e^z\end{aligned}[/tex]

When we substitute our partial derivative values, we get:

[tex]\begin{aligned}Curl F&=\left(z e^x-0\right) \hat{1}-\left(y z e^x-7 x y e^z\right) \hat{\jmath}+\left(0-7 x e^z\right) \hat{k} \\&=z e^x \hat{\imath}-\left(y z e^x-7 x y e^z\right) \hat{\jmath}-7 x e^z \hat{k}\end{aligned}[/tex]

So, we've discovered the curl of the given vector field.

Therefore, the curl of the vector field is:

[tex]=z e^x \hat{\imath}-\left(y z e^x-7 x y e^z\right) \hat{\jmath}-7 x e^z \hat{k}[/tex]

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The correct question is given below:
Consider the following vector field. f(x, y, z) = 7xyezi + yzexk (a) find the curl of the vector field.

The value of term a11 is 21k - 43.

The difference between any two consecutive integers in an arithmetic progression (AP) sequence of numbers is always the same value.

The formula of Arithmetic progression is

a(n) = a1 + (n - 1)d

given that the term of arithmetic sequence and common difference,

a1 = k+7 and d = 2k-5.

now find the value of term a11.

a(n) = a1 + (n - 1)d

now, substitute a(n) = a11, a1 = k+7, d = 2k-5.

a11 = k+7 + (11 - 1)(2k-5)

a11 = k + 7 + 10(2k-5)

a11 = k +7+ 20k - 50

a11 = 21k - 43

Hence,The value of term a11 is 21k - 43.

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Using a system of equations, the prices are given as follows:

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

For this problem, the variables are given as follows:

3 reference and 2 exercise cost RM63, hence:

3x + 2y = 63.

5 reference and 3 exercise cost RM102, hence:

5x + 3y = 102.

Multiplying the first equation by 3 and the second by -2, we have that:

Adding them:

-x = -15 -> x = 15.

Hence:

3(15) + 2y = 63

2y = 18.

y = 9.

Hence the costs are:

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By applying the vertical angles theorem to the two (2) angles, the measures of the labeled angles are equal to 108°.

The vertical angles theorem is also referred to as vertically opposite angles theorem and it states that two (2) opposite vertical angles that are generally formed whenever two (2) lines intersect each other are always congruent i.e equal to each other.

By applying the vertical angles theorem to the two (2) angles, we have:

(x + 81)° = 4x°

x + 81° = 4x°

4x° - x = 81°

3x° = 81°

x° = 81°/3

x° = 27°

Now, we can determine the measures of the labeled angles:

x + 81° = 27° + 81°

x + 81° = 108°

4x° = 4(27)

4x° = 108°

In conclusion, the measures of the labeled angles are equal to 108°.

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Answer:

17 paintings are displayed

Step-by-step explanation:

The circumference of the circle is approximately 17.6 cm  and the area is approximately 24.6 cm²

The perimeter of a circle or an ellipse in geometry is known as the circumference. That is, if the circle were opened up and straightened out to a line segment, the circumference would be the length of the arc. The curve length around any closed figure is more often referred to as the perimeter. The term "circumference" can also refer to the circle's actual location, which corresponds to a disk's edge.

The circumference of a circle is calculated by the formula 2πr .

The given coordinates are A(2,3) and B(4,1)

Here A is the center of the circle and B is any point on the circumference.

Now to calculate the radius we will use the distance formula:

therefore radius [tex]AB=\sqrt{(3-1)^2+(2-4)^2}[/tex]

[tex]or,AB=\sqrt{4+4}\\ \\or, AB=2.828[/tex]

[tex]or, AB\approx2.8[/tex]

Now let us calculate the circumference of the circle:

Circumference = 2πr

or, circumference = 2 × π × 2.8 = 17.592

or, circumference ≈ 17.6 cm

Area = π r²

or, Area = π (2.8)²

or, Area = 24.630 cm²

or, Area ≈ 24.6 cm²

The circumference of the circle is approximately 17.6 cm and the area is approximately 24.6 cm².

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For numbers 5 and 45 the value of the geometric mean of these numbers is

Gm (5, 45) = 15

What is the geometric mean?

  The geometric mean is defined as a type of measure that is calculated as the root of a product between a set of numbers, it can be expressed as follows:

Gm (a, b) = √ab

Then in our case that we have the values 40 and 15, these are equivalent to the variables a and b

a = 5

b = 45

We substitute and solve for the square root

Gm (5, 45) = √5×45

Gm (5, 45) = √225  ;    In this case the square root is direct-

Gm (5, 45) = 15

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The amount that you earned last month would be the total of $17.25

We have the following data

online pay bill per month = 50

The monthly online deposit = 50

The points that are made per car payment = 100

The point that is made per credit card = 1.5 points

The monthly credit card charges of $1,400

The monthly point would be 1400 * 1.5 = $2100

We have to solve for the total points

= 50 + 50 + 100 + 2100

= 2300

We have  $75 per 10,000

For 2300 points, this would be 75/10000 * 2300

= $17.25                                          

Hence the amount that is made in dollars from the points in the last month is $17.25

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(a) A = 1/2 (bh)

(b) b = P/R

(c) 8?

The numeric values for the given function are as follows:

a) f(2.7) = 10.

b) f(1.9) = 7.

c) f(-3.6) = -14.4.

To find the numeric value of a function, we replace each instance of the variable by the desired value.

For this problem, the function is defined as follows:

f(x) = int(4x).

The function int(x) has the output as the largest integer that is less or equal to the value of x.

Hence the numeric values are given by:

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a. The next term in the sequence is 303.75.

b. The sequence is a geometric sequence.

c. The recursive function for the sequence is: aₙ = aₙ₋₁ (1.5) for n≥2.

Given the sequence is : 60,90,135,202.5,....

a. the next term in the sequence is:

Find the common ratio by dividing any term in the sequence by the term that comes before it:

a₂/a₁ = 90/60 = 1.5

a₃/a₂ = 135/90 = 1.5

a₄/a₃ = 202.5/135 = 1.5

The common ratio (r) of the sequence is constant and equals the quotient of two consecutive terms.

r = 1.5

find the sum:

To find the sum of the series, plug the first term: a = 60, the common ratio(r): 1.5 and the number of elements n = 4 into the geometric series sum formula:

Sₙ = a(1₋rⁿ/1₋r)

Sₙ = 60(1 ₋ 1.5⁴/1₋1.5)

Sₙ = 60(1 ₋ 5.0625/1₋1.5)

S₄ = 60(8.125)

S₄ = 478.5

To find the general form of the series, plug the first term:

aₙ = 60×1.5ⁿ⁻¹

use the general form to find the next term:

a₁ = 60

a₂ = a₁rⁿ⁻¹ = 60(1.5)¹ = 90

a₃ = a₁rⁿ⁻¹ = 60(1.5)² = 135

a₄ = a₁rⁿ⁻¹ = 60(1.5)³ = 202.5

a₄ = a₁rⁿ⁻¹ = 60(1.5)⁴ = 303.75

hence the required next term is 303.75

(b) The above sequence is geometric sequence.

(c) recursive function for the sequence : aₙ = aₙ₋₁ (1.5) for n≥2.

Hence we get the required answers.

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Answer:

105.

Step-by-step explanation:

Total score for 10 students = 50 * 10 = 500.

When 1 student joins:

mean score = (500 + x) / 11 = 55    where x is the test score of the new student.

500 + x = 55 * 11

x  = 55*11 - 500

  =  605 - 500

   = 105.

In the given case , we can conclude The test score of the student who joined the class is 105 marks.

Let's solve this problem step by step:

The mean test score of the first 10 students is 50 marks. Therefore, the total marks scored by these 10 students is 10 x 50 = 500 marks.

When the new student joins, the mean score of all 11 students becomes 55 marks.

Therefore, the total marks scored by all 11 students is 11 x 55 = 605 marks.

To find the test score of the new student, we subtract the total marks of the first 10 students from the total marks of all 11 students: 605 - 500 = 105 marks.

Therefore, the test score of the student who joined the class is 105 marks.

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Answer:

the answer is 5.

Step-by-step explanation:

used math-way to solve

Answer:

41. 12, 14

42. 30, 36

43. 0, 1

44. -24, -28

45. 2.5 , 4

46.

47.

48.

49. 63 , 189

Step-by-step explanation:

I don't know answers to 46 , 47 , 48

please make my answer as brainelist

Answer:

Hello,

Step-by-step explanation:

41)

[tex]2,4,6,8,10,...\\\\u_0=2 , u_{n+1}=u_n+2\\\\u_n=2*n+2\\\\[/tex]

42)

[tex]0,6,12,18,24,...\\\\u_0=0 , u_{n+1}=u_n+6\\\\u_n=6*n+0\\\\[/tex]

43)

[tex]-5,-4,-3,-2,-1,...\\\\u_0=-5 , u_{n+1}=u_n+1\\\\u_n=1*n-5=n-5\\\\[/tex]

44)

[tex]-4,-8,-12,-16,-20,...\\\\u_0=-4 , u_{n+1}=u_n-4\\\\u_n=-4*n-4\\\\[/tex]

45)

[tex]-5,-3.5,-2,-0.5,1,...\\\\u_0=-5 , u_{n+1}=u_n+1.5\\\\u_n=1.5*n-5\\\\[/tex]

46)

[tex]-32,-20,-8,4,16,...\\\\u_0=-32 , u_{n+1}=u_n+8\\\\u_n=8*n-32\\\\[/tex]

48)

[tex]0,\dfrac{1}{8},\dfrac{2}{8},\dfrac{3}{8},...\\\\u_0=0 , u_{n+1}=u_n+\dfrac{1}{8}\\\\u_n=\dfrac{1}{8}*n+0\\\\[/tex]

49)

[tex]27,15,3,-9,-21,...\\\\u_0=27 , u_{n+1}=u_n-12\\\\u_n=-12*n+27\\\\[/tex]

47) be carefull this is not a arithmetic sequence but a quadratic.

I show you the general method.

[tex]\begin{array}{c|c|c|c}x&\Delta_1&\Delta_2&\Delta_3\\--&--&--&--\\1&.&.&.\\&&&\\\dfrac{11}{3} &\dfrac{8}{3} &.&.\\&&&\\4 &\dfrac{1}{3}&\dfrac{-7}{3} &.\\&&&\\2 &-2 &\dfrac{-7}{3}&0\\&&&\\--&--&--&--\\\end {array} \\\\Since\ \Delta_3\ is\ null,\ u_n=a*n^2+b*n+c\ has\ for\ degree\ 2.\\[/tex]

[tex]\begin {array} {c|c|c|c|c}n&u_n&equation&\\--&--&-----&\\0&1&=a*0+b*0+c &c=1\\1&\dfrac{11}{3}&=a*1+b*1+c&a+b=\dfrac{8}{3} \\2&4&=a*4+b*2+c&4a+2b=3\\\end {array}\\\\\\\left\{\begin {array} {ccc}4a+2b&=&3\\a+b&=&\dfrac{8}{3}\\\end {array} \right.\\\\\\\left\{\begin {array} {ccc}a&=&\dfrac{-7}{6}\\\\b&=&\dfrac{23}{6}\\\\c&=&1\\\\\end {array} \right.\\\\\\\boxed{u_n=\dfrac{-7}{6}*n^2+\dfrac{23}{6}*n+1} \\[/tex]

[tex]u_n=\dfrac{-7}{6}*n^2+\dfrac{23}{6}*n+1}\\\\u_{n+1}=\dfrac{-7}{6}*(n+1)^2+\dfrac{23}{6}*(n+1)+1}\\\\u_{n+2}=\dfrac{-7}{6}*(n+2)^2+\dfrac{23}{6}*(n+2)+1}\\\\\Delta_1(n)=u_{n+1}-u_{n}=\dfrac{-7}{6}*(2n+1)+\dfrac{23}{6}\\\Delta_1(n+1)=u_{n+2}-u_{n+1}=\dfrac{-7}{6}*(2n+3)+\dfrac{23}{6}\\\\\Delta_2(n)=\Delta_1(n+1)-\Delta_1(n)=\dfrac{-7}{6}*2 \\[/tex]

[tex]\left\{\begin{array}{ccc}u_{n+2}-2u_{n+1}+u_n&=&\dfrac{-7}{3} \\\\u_0&=&1\\\\u_1&=&\dfrac{11}{3} \end {array} \right.[/tex]

The sample space for the events are given as follows -

Picking up a one ball randomly from six balls -

S [1] = {A, B, C, D, E , F}

Picking up a one ball randomly from balls D to F -

S [2] = {D, E , F}

Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur.

Given in the question a bag has six balls labeled A, B, C, D, E , F and one ball is randomly picked.

Sample space is defined as the set of all the outcomes for a given event. In the question given, the event is picking up of a ball randomly. A total number of six balls are there in the bag. Therefore, if a ball is picked up randomly, it could be any of the six. Therefore - the sample space for this event will be -

S [1] = {A, B, C, D, E , F}

Now, for the event of choosing a letter from D to F, we have total three outcomes. So in this case the sample space will be -

S [2] = {D, E , F}

Therefore, the sample space for the events are given as follows -

Picking up a one ball randomly from six balls -

S [1] = {A, B, C, D, E , F}

Picking up a one ball randomly from balls D to F -

S [2] = {D, E , F}

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The amount that would be saved if 7 bags of grapes were bought on Monday is $2 (option J).

The total cost of items bought on Monday and Wednesday would be added together. This amount would be subtracted from the total cost if the items were all bought on Monday.

The first step is to determine the total cost of the grapes bought on Monday and Wednesday.

Total cost of grapes = cost of grapes bought on Monday + cost of grapes bought on Wednesday

Cost of grapes bought on Monday = number of bags bought x price per bag

3 x $3 = $9

Cost of grapes bought on Wednesday = number of bags bought x price per bag

4 X $2.80 = $11.20

Total cost = $9 + $11.20 = $20.20

Cost or the 7 bags of grapes purchased in 1 trip = number of bags bought x price per bag

7 x $2.60 = 18.20

Amount saved = $20.20 - $18.20 = $2

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Explanation:

You want to identify the step that is occurring on each line of the solution to the equation 2(8u+2) = 3(2u-7).

1. 2(8u+2) = 3(2u-7) . . . . Original Equation

2. 16u +4 = 6u -21 . . . . Apply the distributive property

3. 16u +4 -6u = 6u -21 -6u . . . . subtract 6u from both sides

4. 10u +4 = -21 . . . . collect terms

5. 10u -4 = -21 -4 . . . . subtract 4 from both sides

6. 10u = -25 . . . . collect terms

7. 10u/10 = -25/10 . . . . divide both sides by 10

8. u = -2.5 . . . . simplify

The dimension that give the maximum volume will be the sides.

The volume will be 324cm³.

It should be noted that the maximum value is when the rectangle is a square.

Let s be the side of the square.

The perimeter will be 4s.

Solving for s:

= 36/4

= 9cm

It should be noted that area = s²

where s = side

= 9² = 81cm

Therefore, the volume will be

= Area × Side

= 81 × 4

= 324cm³

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SSS postulate of congruence, to prove ∠ L N M ≅ ∠ L N P.

What do you mean by congruent?

In triangles MLN and PLN,

MN ≅ PN,

LM ≅ LP

To prove :

Δ MLN ≅ Δ PLN

  Statement                                              Reason

1. LN ≅ LN                                           1. Common segment

2. MN ≅ PN;                                        2. Given

LM ≅ LP

3. Δ MLN ≅ Δ PLN                              3. SSS  postulate of congruence,

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[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$5000\\ r=rate\to 4.5\%\to \frac{4.5}{100}\dotfill &0.045\\ t=years\dotfill &6 \end{cases} \\\\\\ I = (5000)(0.045)(6) \implies I = 1350[/tex]

The sum of the series of three numbers, which includes the counting number, n, the number before n, and the number after n is 3•n

A series of a sequence of numbers is the sum of the numbers up to a specified term.

The given parameters;

n = A counting number

Counting numbers are the natural numbers which is the set of all positive integers.

Given that n is a counting number, we have that n is a positive whole number.

Which gives;

The whole number before n = n - 1

The whole number after n = n + 1

Which gives;

The sum of n, the whole number before n, and the whole number after n is part of a series and can be written mathematically as follows;

The sum = n + n - 1 + n + 1 = 3•n -1 + 1 = 3•n

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