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lora16 [44]
2 years ago
11

Find X. See image below.

Mathematics
2 answers:
larisa86 [58]2 years ago
7 0

Answer:

x = 9.8 (rounded)

Step-by-step explanation:

the diagonal within the quadrilateral measures

\sqrt{(12)^{2} +(12)^{2} } =\sqrt{288} =16.97

Calculate x on the right triangle of which it is a leg

tan60^{0} =\frac{16.97}{x}

x=\frac{16.97}{tan60^{0} } =9.797

Hope this helps

Nitella [24]2 years ago
4 0

written answer in the picture

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Which statement describes the inverse of m(x) = x^2 – 17x?
DochEvi [55]

Given:

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m(x)=x^2-17x

To find:

The inverse of the given function.

Solution:

We have,

m(x)=x^2-17x

Substitute m(x)=y.

y=x^2-17x

Interchange x and y.

x=y^2-17y

Add square of half of coefficient of y , i.e., \left(\dfrac{-17}{2}\right)^2 on both sides,

x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2        [\because (a-b)^2=a^2-2ab+b^2]

Taking square root on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}

Add \dfrac{17}{2} on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y

Substitute y=m^{-1}(x).

m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}

We know that, negative term inside the root is not real number. So,

x+\left(\dfrac{17}{2}\right)^2\geq 0

x\geq -\left(\dfrac{17}{2}\right)^2

Therefore, the restricted domain is x\geq -\left(\dfrac{17}{2}\right)^2 and the inverse function is m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}.

Hence, option D is correct.

Note: In all the options square of \dfrac{17}{2} is missing in restricted domain.

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