what is the solution to the system of equations below?
3x+y+2z=8
8y+6z=36
12y+2z=12

A. x=2, y=1, z=6
B. x=3, y=0, z=4
C. x=2, y=0, z=6
D. x=3, y=1, z=4

Answer:

Step-by-step explanation:

Answer:

a. y = 2(x + 3)² - 17

b. y = 4(x + 6)² - 154

Step-by-step explanation:

a. y = 2x² + 12x + 1

y = 2[(x² + 6x)] + 1

y = 2[(x + 3)² - 9] + 1

y = 2(x + 3)² - 18 + 1

y = 2(x + 3)² - 17

b. y = 4x² + 48x - 10

y = 4[(x² + 12x)] - 10

y = 4[(x + 6)² - 36)] - 10

y = 4(x + 6)² - 144 - 10

y = 4(x + 6)² - 154

Using the unit rate, the missing values that completes the table are:

A = 5; B = 15; C = 40

Unit rate (m) = change in y/change in x.

5 minutes for 8 Blocks (5, 8) and 10 minutes for 16 blocks (10, 16)are given.

Unit rate (m) = (16 - 8)/(10 - 5) = 8/5

An equation that will define the function is, y = 8/5x. Use it to complete the table.

Find A (y) when x is 5:

y = 8/5(5) = 8

The value of A is: 5

Find B (x) when y is 24:

24 = 8/5(x)

5(24) = 8x

120 = 8x

120/8 = x

15 = x

The value of B is: 15

Find A (y) when x is 25:

y = 8/5(25) = 40

The value of C is: 40

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Answer:

Range: (-10 , 14)

Step-by-step explanation:

Given information:

Range: (x , y)?

Plug in domain of x = -4 and x = 4 into equation to find range.

f(-4) = 3 * -4 + 2 = -10

f(4) = 12 + 2 = 14

Range: (-10 , 14)

The area of the triangle will be 24912 sq. units. Square units and other similar units are used to measure area.

The space filled by a flat form or the surface of an item is known as the area.

The number of unit squares that cover the surface of a closed-form is the figure's area.

For:

(X1, Y1) = (1, 15)

(X2, Y2) = (-2, 1)

d = 14.317821

For:

(X₂, Y₂) = (-2, 1)

(X₃, Y₃) = (4, 5)

d = 7.211103

For applying the pythogorous them we need the right angle triangle obtained by bisect from the mid point.

The value of the base is;

⇒7.2 / 2

⇒3.6

apply the pythogorous theorem for finding the height;

h² = p² + b²

14.31² = p² + 3.6²

p = 13.84

The area of the triangle is;

[tex]\rm A = \frac{1}{2}\times b \times h \\\\ A= \frac{1}{2} \times 3.6 \times 13.84 \\\\ A = 24.912[/tex]

Hence, the area of the triangle will be 24912 sq. units.

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Step-by-step explanation:

the slope is the factor of x in an equation

y = ax + b

we have here

3y = -5x + 21

to get to the general format above we need to divide everything by 3 :

y = -5/3 x + 7

so, we see, the slope is -5/3.

the perpendicular (angle of 90°) slope is the original slope turned upside-down and with flipped sign :

3/5

so, I guess the correct answer option is c.

but it is not clear what you wrote there, as there is a "-" sign somehow in all 4 answers.

Answer:

x = 8

Step-by-step explanation:

[tex]\sf If\:\: \overline{NO} \parallel \overline{KJ}\:\:then\:\: \triangle LNO \sim\triangle LKJ[/tex]

Therefore:

[tex]\implies \sf \overline{LN} : \overline{LO} = \overline{LK} : \overline{LJ}[/tex]

[tex]\implies (x-3):(x-4)=(x-3)+(x+2):(x-4)+x[/tex]

[tex]\implies \dfrac{x-3}{x-4}=\dfrac{2x-1}{2x-4}[/tex]

[tex]\implies (x-3)(2x-4)=(2x-1)(x-4)[/tex]

[tex]\implies 2x^2-10x+12=2x^2-9x+4[/tex]

[tex]\implies -10x+12=-9x+4[/tex]

[tex]\implies 12=x+4[/tex]

[tex]\implies x=8[/tex]

Answer:

40 45 because the minimum internal angle is 60

Answer:

X=62 degrees

Step-by-step explanation:

The solution is in the image

Answer:

62°

Step-by-step explanation:

We know that the sum of the interior angles in a triangle is added up to 180°.

Therefore,

68.5° + 49.5° + x = 180°

118° + x° = 180°

x = 180° - 118°

x = 62°

If the value of the z-score is 1. Then the probability that a cat will weigh less than 11 pounds will be 0.84134.

The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.

The z-score is given as

z = (x - μ) / σ

Where μ is the mean, σ is the standard deviation, and x is the sample.

The weight of a cat is normally distributed with a mean of 9 pounds and a standard deviation of 2 pounds.

Then the probability that a cat will weigh less than 11 pounds will be

The value of z-score will be

z = (11 – 9) / 2

z = 1

Then the probability will be

P(x < 11) = P(z < 1)

P(x < 11) = 0.84134

Thus, the probability that a cat will weigh less than 11 pounds will be 0.84134.

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Answer:

B. shifting

Step-by-step explanation:

When the price is rising or shifting higher, the demand curve moves to the left. But, t here is a corresponding change in the demand curve in that scenario while the cost stays unchanged. Since it relies on variables other than price, the simultaneous shift might go either left or right. However, the upward-pointing line depicts a rise in price and a fall in supply. 

The ordered pair (7,2) is viable and Option B is the correct answer.

Relationship between variables defines the way one variable is dependent upon the other variable.

It is given that x is the number of visitors and y is the number of winners,

It has to be seen and chosen that which ordered pair makes sense

The ordered pair is viable if the no. of visitor is positive and more than the number of winners.

Therefore ordered pair (7,2) is viable and Option B is the correct answer.

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An equation is formed of two equal expressions. The estimated solution of the two system of equations is at (5/2,4/3). Thus, the correct option is D.

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

The solution of the system of equation is the point at which the two lines will intersect as shown below. Therefore, the solution will be,

Solution = (5/2, 4/3)

Hence, the estimated solution of the two system of equations is at (5/2,4/3). Thus, the correct option is D.

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Answer:

a

Step-by-step explanation:

given a quadratic equation in standard form

ax² + bx + c = 0 ( a ≠ 0 )

then the discriminant

Δ = b² - 4ac

• if b² - 4ac > 0 then 2 real solutions

• if b² - 4ac = 0 then 2 real and equal solutions

• if b² - 4ac < 0 then no real solutions

given

[tex]\frac{3}{4}[/tex] x² - 3x = - 4 ( add 4 to both sides )

[tex]\frac{3}{4}[/tex] x² - 3x + 4 = 0 ← in standard form

with a = [tex]\frac{3}{4}[/tex] b = - 3 , c = 4

then

b² - 4ac = (- 3)² - ( 4 × [tex]\frac{3}{4}[/tex] × 4) = 9 - 12 = - 3

since b² - 4ac < 0 then equation has no real solutions

Answer:

a.  -3; no real solutions.

Step-by-step explanation:

Discriminant

[tex]\boxed{b^2-4ac }\quad\textsf{when}\:ax^2+bx+c=0[/tex]

[tex]\textsf{When }\:b^2-4ac > 0 \implies \textsf{two real solutions}.[/tex]

[tex]\textsf{When }\:b^2-4ac=0 \implies \textsf{one real solution}.[/tex]

[tex]\textsf{When }\:b^2-4ac < 0 \implies \textsf{no real solutions}.[/tex]

Given equation:

[tex]\dfrac{3}{4}x^2-3x=-4[/tex]

Add 4 to both sides of the equation so that it is in standard form:

[tex]\implies \dfrac{3}{4}x^2-3x+4=-4+4[/tex]

[tex]\implies \dfrac{3}{4}x^2-3x+4=0[/tex]

Therefore, the variables are:

[tex]a=\dfrac{3}{4}, \quad b=-3, \quad c=4[/tex]

Substitute these values into the discriminant formula to find the value of the discriminant:

[tex]\begin{aligned}\implies b^2-4ac&=(-3)^2-4\left(\dfrac{3}{4}\right)(4)\\&=9-(3)(4)\\&=9-12\\&=-3\\\end{aligned}[/tex]

Therefore, as -3 < 0, the discriminant is less than zero.

This means there are no real solutions.

Answer:

x=24, y=4

Step-by-step explanation:

x=6y

1/x+1/y=7/24,

1/6y+1/y=7/24

1/6y+6/6y=7/24

(1+6)/6y=7/24

7/6y=7/24, then

6y=24

y=24/6

y=4

x=6y=6*4=24

Answer:

well not sure but I think it is 8° or 17.8°

Answer:

1630

Step-by-step explanation:

In words you are looking for 6% of what number is 97.80, turn that into an Algebra equation .06x = 97.80 so x = 97.80/.06 so x = 1630

Answer:

C

Step-by-step explanation:

Answer C is correct.  The absolute value of 10 is 10 and that of -10 is 10.  Same result.

Answer:

[tex]28{cm}^{2} [/tex]

Step-by-step explanation:

we know

area of rectangle=l*b=(7*4)sq cm=28sq cm

Based on the key features, the end behavior and the zeros of function h [h(x) = -x³ - 9x² +4x +96], the statement that is true of the above graph is: "The graph is positive on the intervals (-8, -4) and (3, infinity) (Option B)"

A function's "end behavior" refers to how the function's graph behaves at its "ends" on the x-axis.

In other words, if we look at the right end of the x-axis (as x approaches + ∞) and the left end of the x-axis (as x approaches - ∞ ), the end behavior of a function represents the trend of the graph.

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The error from Jorge's arithmetic operation on the given algebraic expression is that he distributed it incorrectly.

An algebraic expression is a mathematical equation that is made up of variables together with arithmetic operations.

From the given expression, we have:

4x - x + 2 + 6 = 6x + 16

Add similar elements together;

3x + 8 = 6x + 16

Using distributive property, subtract 8 from both sides:

3x + 8 - 8 = 6x + 16 - 8

3x = 6x + 8

Simplify

3x-6x = 8

-3x = 8

Divide both sides by -3

-3/-3x = -8/3

x =  -8/3

So from Jorge's calculation, because he distributed incorrectly, we can conclude that could be his error.

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Answer: i think no?

Step-by-step explanation:

AD and BE are both parallel lines (they are parallel to eachother), so they will never intersect

CF and GF intersect at point F (i think)

Bg and Ac intersect at point B (i think)

i dont want to give a definite answer in the event im wrong bc I just learned this like a few weeks ago-

The correct value that equates to this expression is 12‐³². Letter A

.

To solve this expression, just: eliminate the parentheses and multiply the exponents among themselves;[tex] \boxed{ \large \sf (a {}^{n} ) {}^{m} \rightarrow a {}^{n \times m} } \\ \\ [/tex]

[tex]{ = \large \sf (12{}^{-4} ) {}^{8} } [/tex]

[tex]{ = \large \sf 12{}^{-4 \times 8} } [/tex]

[tex] \pink{ \boxed{ = \large \sf 12{}^{-32} } } \\ [/tex]

Therefore, the answer will be 12‐³²

Answer:

Step-by-step explanation:

A function can only have one output for an input.  That is, for any value of x, there must be a unique value of y.

{(-6,-1), (13,8), (1,6), (1,-10)}   Not a  Function:  (1,6) and (1,-10)

{(10,5), (10,-5), (5,10), (5,-10)}  Not a  Function:  (10,5) and (10,-5)

{(3,5), (-17,-5), (3,-5), (-17,5)}   Not a Function:  (3,5) and (3,-5)

{(10,5), (-10,-5), (5,10), (-5,-10)}  Function:  No duplicate values of y for a value of x.

Answer:

B

Step-by-step explanation:

-2|x - 1| = -4

|x - 1| = 2

since we are dealing with a function that brings 2 values to the same result, the reverse function (needed to find the values of x that create the result y) has 2 branches :

(x - 1) = 2

and

(x - 1) = -2

x - 1 = 2

x = 3

x - 1 = -2

x = -1

therefore, B is the right answer.

To find the maxima and minima of the function, we need to calculate the derivative of the function. Note, before the denominator is a perfect square trinomial, so the function can be simplified as

[tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle f(x) = \frac{x^2 - x - 2}{(x - 3)^2}} \end{gathered}$}[/tex]

So the derivative is:

  [tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle f'(x) = \frac{(2x - 1)(x - 3)^2 - 2(x - 3)(x^2 - x - 2)}{(x - 3)^4} } \end{gathered}$}[/tex]

Simplifying the numerator, we get:

                 [tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle f'(x) = \frac{(x - 3)(-5x + 7)}{(x - 3)^4} = \frac{-5x + 7}{(x - 3)^3} } \end{gathered}$}[/tex]

The function will have a maximum or minimum when f'(x) = 0, that is,

                  [tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle f'(x) = \frac{-5x + 7}{(x - 3)^3} = 0 } \end{gathered}$}[/tex]

which is true if -5x + 7 = 0. Then x = 7/5.

To determine whether x = 7/5 is a maximum, we can use the second derivative test or the first derivative test. In this case, it is easier to use the first derivative test to avoid calculating the second derivative. For this, we evaluate f'(x) at a point to the left of x = 7/5 and at a point to the right of it (as long as it is not greater than 3). Since 1 is to the left of 7/5, we evaluate:

                    [tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle f(1) = \frac{-5 + 7}{(1 - 3)^3} = \frac{2}{-8} < 0} \end{gathered}$}[/tex]

Likewise, since 2 is to the right of 7/5, then we evaluate:

                                   [tex]\large\displaystyle\text{$\begin{gathered}\sf \displaystyle \bf{\frac{-10 + 7}{(2 - 3)^3} = \frac{-3}{-1} > 0} \end{gathered}$}[/tex]

Note that to the left of 7/5 the derivative is negative (the function decreases) and to the right of 7/5 the derivative is positive (the function increases).

The value of f(x) at 7/5 is:

                               [tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle f\left(\tfrac{7}{5}\right) = \frac{\tfrac{49}{25} - \tfrac{7}{5} - 2}{\tfrac{49}{25} - 6 \cdot \tfrac{7}{5} + 9} = -\frac{9}{16} } \end{gathered}$}[/tex]

This means that [tex]\bf{\left( \frac{7}{5}, -\frac{9}{16} \right)}[/tex] is a minimum (and the only extreme value of f(x)).

[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}[/tex]

Answer:

[tex]\text{Minimum at }\left(\dfrac{7}{5},-\dfrac{9}{16}\right)[/tex]

Step-by-step explanation:

The local maximum and minimum points of a function are stationary points (turning points).  Stationary points occur when the gradient of the function is zero.  Differentiation is an algebraic process that finds the gradient of a curve.

To find the stationary points of a function:

[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Quotient Rule for Differentiation}\\\\If $y=\dfrac{u}{v}$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{v \dfrac{\text{d}u}{\text{d}x}-u\dfrac{\text{d}v}{\text{d}x}}{v^2}$\\\end{minipage}}[/tex]

[tex]\text{Given function}: \quad \text{f}(x)=\dfrac{x^2-x-2}{x^2-6x+9}[/tex]

Differentiate the function using the Quotient Rule:

[tex]\text{Let }u=x^2-x-2 \implies \dfrac{\text{d}u}{\text{d}x}=2x-1[/tex]

[tex]\text{Let }v=x^2-6x+9 \implies \dfrac{\text{d}v}{\text{d}x}=2x-6[/tex]

[tex]\begin{aligned}\implies \dfrac{\text{d}y}{\text{d}x} & =\dfrac{(x^2-6x+9)(2x-1)-(x^2-x-2)(2x-6)}{(x^2-6x+9)^2}\\\\& =\dfrac{(2x^3-13x^2+24x-9)-(2x^3-8x^2+2x+12)}{(x^2-6x+9)^2}\\\\\implies \text{f}\:'(x)& =\dfrac{-5x^2+22x-21}{(x^2-6x+9)^2}\\\\\end{aligned}[/tex]

Set the differentiated function to zero and solve for x:

[tex]\begin{aligned}\implies \text{f}\:'(x)& =0\\\\\implies \dfrac{-5x^2+22x-21}{(x^2-6x+9)^2} & = 0\\\\-5x^2+22x-21 & = 0\\\\-(5x-7)(x-3) & = 0\\\\\implies 5x-7 & = 0 \implies x=\dfrac{7}{5}\\\\\implies x-3 & = 0 \implies x=3\end{aligned}[/tex]

Put the x-values back into the original equation to find the y-values:

[tex]\implies \text{f}\left(\frac{7}{5}\right)=\dfrac{\left(\frac{7}{5}\right)^2-\left(\frac{7}{5}\right)-2}{\left(\frac{7}{5}\right)^2-6\left(\frac{7}{5}\right)+9}=-\dfrac{9}{16}[/tex]

[tex]\implies \text{f}(3)=\dfrac{\left(3\right)^2-\left(3\right)-2}{\left(3\right)^2-6\left(3\right)+9}=\dfrac{4}{0} \implies \text{unde}\text{fined}[/tex]

Therefore, there is a stationary point at:

[tex]\left(\dfrac{7}{5},-\dfrac{9}{16}\right)\:\text{only}[/tex]

To determine if it's a minimum or a maximum, find the second derivative of the function then input the x-value of the stationary point.

Differentiate f'(x) using the Quotient Rule:

Simplify f'(x) before differentiating:

[tex]\begin{aligned}\text{f}\:'(x) & =\dfrac{-5x^2+22x-21}{(x^2-6x+9)^2}\\\\& = \dfrac{-(5x-7)(x-3)}{\left((x-3)^2\right)^2}\\\\& = \dfrac{-(5x-7)(x-3)}{(x-3)^4}\\\\& = -\dfrac{(5x-7)}{(x-3)^3}\\\\\end{aligned}[/tex]

[tex]\text{Let }u=-(5x-7) \implies \dfrac{\text{d}u}{\text{d}x}=-5[/tex]

[tex]\text{Let }v=(x-3)^3 \implies \dfrac{\text{d}v}{\text{d}x}=3(x-3)^2[/tex]

[tex]\begin{aligned}\implies \dfrac{\text{d}^2y}{\text{d}x^2} & =\dfrac{-5(x-3)^3+3(5x-7)(x-3)^2}{(x-3)^6}\\\\& =\dfrac{-5(x-3)+3(5x-7)}{(x-3)^4}\\\\\implies \text{f}\:''(x)& =\dfrac{10x-6}{(x-3)^4}\end{aligned}[/tex]

Therefore:

[tex]\text{f}\:''\left(\dfrac{7}{5}\right)=\dfrac{625}{512} > 0 \implies \text{minimum}[/tex]

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \: y= - \cfrac{x}{ 3} + 1 [/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

Equation of line (two point form) :

[tex]\qquad \tt \rightarrow \: (y - y_1) = \cfrac{y _1- y_2}{ x_1 - x_2} (x - x_1)[/tex]

[tex]\qquad \tt \rightarrow \: (y - 2) = \cfrac{2 - ( - 1)}{ - 3 - 6} (x - ( - 3))[/tex]

[tex]\qquad \tt \rightarrow \: (y - 2) = \cfrac{2 + 1}{ - 9} (x + 3)[/tex]

[tex]\qquad \tt \rightarrow \: (y - 2) = - \cfrac{3}{ 9} (x + 3)[/tex]

[tex]\qquad \tt \rightarrow \: (y - 2) = - \cfrac{1}{ 3} (x + 3)[/tex]

[tex]\qquad \tt \rightarrow \: y - 2= - \cfrac{x}{ 3} - \cfrac{3}{3} [/tex]

[tex]\qquad \tt \rightarrow \: y = - \cfrac{x}{ 3} - 1 \cfrac{}{} + 2[/tex]

[tex]\qquad \tt \rightarrow \: y = - \cfrac{x}{ 3} + 1[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Answer:

complete this

Step-by-step explanation:

yeah do it

Answer:

[tex]\fbox {D. 3,200.4 mm}[/tex]

Step-by-step explanation:

Given :

[ 1 inch = 2.54 centimeters ]

Unit conversions to keep in mind :

Solving