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3 years ago

6

Solve the equation 16y(2–y)+(4y–5)2=0

1 answer:

zmey [24]3 years ago

6 0

16y(2−y)+(4y−5)(2)=0

Step 1: Simplify both sides of the equation.

−16y2+40y−10=0

Step 2: Use quadratic formula with a=-16, b=40, c=-10.

y=−b±√b2−4ac2a

y=−(40)±√(40)2−4(−16)(−10)2(−16)

y=−40±√960−32

y=54+−14√15 or y=54+14√15

Answer:

y=54+−14√15 or y=54+14√15

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Answer:

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Step-by-step explanation:

Let <em>w</em> represent the width of the rectangle.

We are given that the perimeter of the rectangle is 20 cm, this means that:

$20=2(x + w)$

Let's put <em>w</em> in terms of <em>x</em>. Divide both sides by two:

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And solve for <em>w:</em>

<em /> $w=10-x$ <em />

So, the rectangle measures <em>x</em> by (10 -<em> x</em>) cm.

<em />

According to the Pythagorean Theorem:

$a^2+b^2=c^2$

<em>a</em> and <em>b</em> are the legs and <em>c</em> is the hypotenuse.

Substitute <em>x</em> for <em>a, w</em> for <em>b</em>, and 8 for <em>c:</em>

<em /> $x^2+w^2=8^2$ <em />

Simplify and substitute:

$x^2+(10-x)^2=64$

Square:

$x^2+(100-20x+x^2)=64$

Isolate the equation. So:

$2x^2-20x+36=0$

Since the leading coefficient is one, divide both sides by two:

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Therefore, <em>a</em> = -10 and <em>b</em> = 18.

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