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2 years ago

The measures of the angles of triangle ABC are given by the expression and the table

Mathematics

2 answers:

n200080 [17]2 years ago

8 0

In a triangle all the angles add up to 180 degrees.So we know that
(6x-1)+20+(x+14)= 180
6x+x+14-1+20=180
v v
7x + 13 + 20= 180
v
7x + 33=180
-33 -33
0 147
7x/7 = 147/7
x = 21
x=21
Then you substitute:
(6x-1) (x+14)
(6×21-1) (21+14)
v v
126-1 35
125 degrees / 35 degrees
125+35+20=180

OlgaM077 [116]2 years ago

8 0

The measurement of angle A=125 degrees
the measurement of angle C= 35 degrees

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

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$m(x)=x^2-17x$

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The inverse of the given function.

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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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