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2 years ago

50 POINTS!!!!!!!

Mathematics

1 answer:

stira [4]2 years ago

3 0

Answer:

For this transition of equations, the graph of g(x) will be translated left 2 units with respect to the graph of f(x), so your answer choice will be A.

Step-by-step explanation:

In this equation, g(x) is changed by adding 2 and closing part of the equation in parenthases, this results in the translation 2 units left, which can be proven by a graph and my answer.

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What property is shown?

svetoff [14.1K]

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2 years ago

Meg skateboards 23 of a mile in 49 of an hour. At this rate, how far will she skateboard in an hour?

kipiarov [429]

Answer:

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Step-by-step explanation:

8 0

2 years ago

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10-5(9n-9) simplify the expressions

eduard

Answer:

−45n+55

Step-by-step explanation:

10-5(9n-9)

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8 0

2 years ago

Given csc(A) = 60/16 and that angle A is in Quadrant I, find the exact value of sec A in simplest radical form using a rational

Katarina [22]

Answer:

$\displaystyle \sec A=\frac{65}{63}$

Step-by-step explanation:

We are given that:

$\displaystyle \csc A=\frac{65}{16}$

Where A is in QI.

And we want to find sec(A).

Recall that cosecant is the ratio of the hypotenuse to the opposite side. So, find the adjacent side using the Pythagorean Theorem:

$a=\sqrt{65^2-16^2}=\sqrt{3969}=63$

So, with respect to A, our adjacent side is 63, our opposite side is 16, and our hypotenuse is 65.

Since A is in QI, all of our trigonometric ratios will be positive.

Secant is the ratio of the hypotenuse to the adjacent. Hence:

$\displaystyle \sec A=\frac{65}{63}$

6 0

2 years ago

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Solve.

Ksenya-84 [330]

Solve.

x² + 4x + 4 = 18

A x=−4±3√ 2

B x=2±3√ 2

C x=4±9√ 2

D x=−2±3√2

 $x^2 + 4x + 4 = 18 \\ \\ x^2+4x+4-18= 0 \\ \\ x^2+4x-14=0 \\ \\ x_1_y_2= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{16-(-56)} }{2} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{72} }{2} \\ \\ \\ x_1= \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2= \dfrac{-4+ \sqrt{72} }{2}\\ \\ \\ x_1= \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2= \dfrac{-4- \sqrt{2^3*3^2} }{2}$

 $x_1= \dfrac{-4+2*3 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 2*3\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4+6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 6\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4}{2} + \dfrac{6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4}{2}- \dfrac{ 6\sqrt{2} }{2} \\ \\ \\ x_1= -2 + 3 \sqrt{2} \qquad\qquad\quad x_2= -2- 3\sqrt{2} \\ \\ \\ \boxed{x= -2\pm 3\sqrt{2} } \to D)$

7 0

3 years ago

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