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2 years ago

Identify the zeros given the quadratic equation y= -16(x+7)(x-5)

Mathematics

2 answers:

Darina [25.2K]2 years ago

7 0

Answer:

wouldn't it be -7 and 5? I'm pretty sure that you switch the signs around.

Step-by-step explanation:

set y=0 and solve. -7 + 7 = 0 and 5 - 5 = 0. my teacher told me to switch the signs for each factored term. hope this helps.

garik1379 [7]2 years ago

7 0

Answer:

x = -7

x = 5

Step-by-step explanation:

Hello!

We can use the Zero Product Property to solve for the Zeroes.

<h3>Zero product Property</h3>

The zero product property stars by factoring a quadratic. You then set each of the factors to zero, and solve for the variable in the factor.

Factored form: y = a(x - h)(x - k)

0 = a(x - h)(x - k)
0 = (x - h)(x - k) Divide by a

Now, let's call each factor, A and B. To get to 0, either A, B, or both have to be 0. We prefer to solve for both.

<h3>Solve</h3>

y = -16(x + 7)(x - 5)
0 = -16(x + 7)(x - 5)
0 = (x + 7)(x - 5) Divide by -16

x + 7 = 0, x = -7
x - 5 = 0, x = 5

The zeroes of the Quadratic is -7 and 5.

Why do we replace it with 0?

Zeroes means the x-intercepts, or the roots of a quadratic. An x-intercet is when y is equal to 0. So we replace y with 0, and solve for x.

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alex41 [277]

Answer: $\frac{8}{3}\times \sqrt{\frac{2}{5}}$

Step-by-step explanation:

Given two upward facing parabolas with equations

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x= $\pm \sqrt{\frac{2}{5}}$

area enclosed by them is given by

A= $\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx$

A= $\int_{\sqrt{-\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left ( 2-5x^2\right )dx$

A= $4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ]$

A= $\frac{8}{3}\times \sqrt{\frac{2}{5}}$

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3 years ago

Which value of x is in the domain of f(x) = square root of x-5

Gelneren [198K]

The answer is A. x= 5

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3 years ago

Please answer and explain how you did it.

alukav5142 [94]

Refer to the attached image.

Given: $m \angle A = 65^\circ$ and measure of exterior angle at C = $117^\circ$ .

We have to determine the measure of angle B and angle BCA.

By applying exterior angle property of the triangle which states:

"An exterior angle of a triangle is equal to the sum of the opposite interior angles".

So, exterior angle C = $m \angle A + m \angle B$

$117^\circ = 65^\circ+ m \angle B$

$m \angle B = 52^\circ$

Now, applying angle sum property in triangle ABC which states:

"The sum of all the angles of a triangle is 180 degrees".

$m \angle A + m \angle B + m \angle BCA = 180^\circ$

$65^\circ + 52^\circ + m \angle BCA = 180^\circ$

$117^\circ + m \angle BCA = 180^\circ$

$m \angle BCA = 180^\circ - 117^\circ$

$m \angle BCA =63^\circ$

Therefore, the measure of $m \angle B = 52^\circ$ and $m \angle BCA =63^\circ$ .

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3 years ago

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tester [92]

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3 years ago

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Ratling [72]

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QUESTION 2: 5xy + 9 = 44

5xy = 35

xy = 7

solution pairs are:

C. (1, 7) and (7, 1)

not mentioned: (-1.-7) and (-7, -1)

Hope this helps

Plzz don't forget to rate and thanks me

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