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antoniya [11.8K]

2 years ago

8

John has 48 cm tiles he wants to use to create a mosaic he wants the mosaic to be rectangular with the length that is 2 cm longe

r than the width which equation could John solve to find W the greatest width in centimeters he can use for the mosaic

1 answer:

bonufazy [111]2 years ago

7 0

I need help my dang self

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Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u –

Verdich [7]

Answer:

The solutions of the original equation are x=-5 and x=-2

Step-by-step explanation:

we have

$(x+3)^2+(x+3)-2=0$

Let

$u=(x+3)$

Rewrite the equation

$(u)^2+(u)-2=0$

Complete the square

$u^2+u=2$

$u^2+u+1/4=2+1/4$

$u^2+u+1/4=9/4$

rewrite as perfect squares

$(u+1/2)^2=9/4$

square root both sides

$(u+1/2)=\pm\frac{3}{2}$

$u=(-1/2)\pm\frac{3}{2}$

$u=(-1/2)+\frac{3}{2}=1$

$u=(-1/2)-\frac{3}{2}=-2$

the solutions are

u=-2,u=1

<em>Alternative Method</em>

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$(u)^2+(u)-2=0$

so

$a=1\\b=1\\c=-2$

substitute in the formula

$u=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}$

$u=\frac{-1\pm\sqrt{9}} {2}$

$u=\frac{-1\pm3} {2}$

$u=\frac{-1+3} {2}=1$

$u=\frac{-1-3} {2}=-2$

the solutions are

u=-2,u=1

<em>Find the solutions of the original equation</em>

For u=-2

$-2=(x+3)$ ----> $x=-2-3=-5$

For u=1

$1=(x+3)$ ----> $x=1-3=-2$

therefore

The solutions of the original equation are

x=-5 and x=-2

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Suppose that y varies directly with x and y=10 when x=-20. What is y when x =15?

USPshnik [31]

Answer:

15/2

Step-by-step explanation:

You want to first find the connection between the first statement, y=10 and x=20. The connection is that Y is X divided by 2. Then you want to just do the same equation to the y and keep it a fraction. (Hint. a fraction is the same as what you do to divide, its just the equation before you find your answer.) so your answer is 15/2.

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