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Sophie [7]
3 years ago
15

The zeros of the function f(x)=(x+2)^2 - 25 are?

Mathematics
2 answers:
zloy xaker [14]3 years ago
5 0
<u>Zeros of the function</u>
f(x) = (x + 2)² - 25
f(x) = (x + 2)(x + 2) - 25
f(x) = x(x + 2) + 2(x + 2) - 25
f(x) = x(x) + x(2) + 2(x) + 2(2) - 25
f(x) = x² + 2x + 2x + 4 - 25
f(x) = x² + 4x + 4 - 25
f(x) = x² + 4x - 21
x² + 4x - 21 = 0
x = <u>-(4) +/- √((4)² - 4(1)(-21))</u>
                      2(1)
x = <u>-4 +/- √(16 + 84)</u>
                   2
x = <u>-4 +/- √(100)
</u>               2<u>
</u>x = <u>-4 +/- 10
</u>            2<u>
</u>x = -2 <u>+</u> 5<u>
</u>x = -2 + 5    x = -2 - 5
x = 3           x = -7
f(x) = x² + 4x - 21
f(3) = (3)² + 4(3) - 21
f(3) = 9 + 12 - 21
f(3) = 21 - 21
f(3) = 0
(x, f(x)) = (3, 0)
or
f(x) = x² + 4x - 21
f(-7) = (-7)² + 4(-7) - 21
f(-7) = 49 - 28 - 21
f(-7) = 21 - 21
f(-7) = 0
(x, f(x)) = (-7, 0)

<u>Vertex</u>
<u>X - Intercept</u>
<u />-b/2a = -(4)/2(1) = -4/2 = -2

<u>Y - Intercept</u>
y = x² + 4x - 21
y = (-2)² + 4(-2) - 21
y = 4 - 8 - 21
y = -4 - 21
y = -25
(x, y) = (-2, -25)
<u />
Serjik [45]3 years ago
5 0
Y=x^2+4x+4-25
y=x^2+4x-21
y=(x+7)(x-3)
zeros are -7 and 3
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Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

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Using translation concepts, it is found that the transformations to create function d are given as follows:

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<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, the parent cosine function is given by:

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