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Katarina [22]
3 years ago
12

If y=-27 when x= -9 find x when y=42

Mathematics
1 answer:
Novay_Z [31]3 years ago
8 0

\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf \textit{we know that } \begin{cases} y=-27\\ x=-9 \end{cases}\implies -27=k(-9)\implies \cfrac{-27}{-9}=k \\\\\\ 3=k\qquad therefore\qquad \boxed{y=3x} \\\\\\ \textit{when y = 42, what is \underline{x}?}\qquad 42=3x\implies \cfrac{42}{3}=x\implies 14=x

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What is the equation of the following line written in general form? (The y-intercept is 7.)
Viktor [21]

Answer:

<h2>3x - y + 7 = 0</h2>

Step-by-step explanation:

The slope-intercept form of an equation of a line:

y=mx+b

m - slope

b - y-intercept

Put the given y-intercept b = 7 and the coordinates of the point (-2, 1) to the equation:

1=-2m+7          <em>subtract 7 from both sides</em>

-6=-2m       <em>divide both sides by (-2)</em>

3=m\to m=3

We have the equation:

y=3x+7

Convert it to the general form Ax+By+C=0:

y=3x+7              <em>subtract 3x and 7 from both sides</em>

-3x+y-7=0           <em>change the signs</em>

3x-y+7=0

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Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge
Ede4ka [16]

Answer:

Part a) The quadratic function is 4x^{2} +20x-39=0

Part b) The value of x is 1.5\ in

Part c) The photo and frame together are 7\ in wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0

Part b) What is the value of x?

Solve the quadratic equation 4x^{2} +20x-39=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem

we have

4x^{2} +20x-39=0

so

a=4\\b=20\\c=-39

substitute in the formula

x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}

x=\frac{-20(+/-)\sqrt{1,024}} {8}

x=\frac{-20(+/-)32} {8}

x=\frac{-20(+)32} {8}=1.5\ in  -----> the solution

x=\frac{-20(-)32} {8}=-6.5\ in

Part c) How wide are the photo and frame together?

(4+2x)=4+2(1.5)=7\ in

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Answer:

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Step-by-step explanation:

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