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3 years ago

Which of the following is a zero for the function f(x) = (x − 15)(x + 1)(x − 10)?

Mathematics

2 answers:

antiseptic1488 [7]3 years ago

5 0

we have the function

$f(x)=(x-15)(x+1)(x-10)$

we know that

The zero's of the function are the values of variable x when the value of the function is equal to zero

in the given function the zero's are

$x=15\\x=-1\\x=10$

therefore

<u>the answer is</u>

$x=15$

MrRissso [65]3 years ago

5 0

The answer should be x=15 because once u subtract 15-15 u get 0 and then u substitute 15 in "x" and then. Multiply 0 by everything else n u get 0:))

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In Triangle XYZ, measure of angle X = 49° , XY = 18°, and

marissa [1.9K]

Answer:

There are two choices for angle Y: $Y \approx 54.987^{\circ}$ for $XZ \approx 15.193$ , $Y \approx 27.008^{\circ}$ for $XZ \approx 8.424$ .

Step-by-step explanation:

There are mistakes in the statement, correct form is now described:

<em>In triangle XYZ, measure of angle X = 49°, XY = 18 and YZ = 14. Find the measure of angle Y:</em>

The line segment XY is opposite to angle Z and the line segment YZ is opposite to angle X. We can determine the length of the line segment XZ by the Law of Cosine:

$YZ^{2} = XZ^{2} + XY^{2} -2\cdot XY\cdot XZ \cdot \cos X$ (1)

If we know that $X = 49^{\circ}$ , $XY = 18$ and $YZ = 14$ , then we have the following second order polynomial:

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$XZ^{2}-23.618\cdot XZ +128 = 0$ (2)

By the Quadratic Formula we have the following result:

$XZ \approx 15.193\,\lor\,XZ \approx 8.424$

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$XZ^{2} = XY^{2} + YZ^{2} - 2\cdot XY \cdot YZ \cdot \cos Y$

$\cos Y = \frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ}$

$Y = \cos ^{-1}\left(\frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ} \right)$

1) $XZ \approx 15.193$

$Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-15.193^{2}}{2\cdot (18)\cdot (14)} \right]$

$Y \approx 54.987^{\circ}$

2) $XZ \approx 8.424$

$Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-8.424^{2}}{2\cdot (18)\cdot (14)} \right]$

$Y \approx 27.008^{\circ}$

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