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4 years ago

8

Select the correct answer.

1 answer:

lora16 [44]4 years ago

6 0

Answer:

A is the inverse function,

Step-by-step explanation:

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Write and equation based on the graph. Please help me 20 points!

zaharov [31]

Answer:

y = 3/5x - 1

Step-by-step explanation:

The slope found from counting the units is 3/5, and the y-intercept is -1, hence 3/5x - 1

8 0

2 years ago

What is the radius for the circle given by the equation?

alisha [4.7K]

Use:

$(a-b)^2=a^2-2ab+b^2\qquad(*)$

$x^2+y^2-6y-12=0\ \ \ \ |+12\\\\x^2+y^2-6y=12\\\\x^2+y^2-2y\cdot3=12\ \ \ \ |+3^2\\\\x^2+\underbrace{y^2-2\cdot y\cdot3+3^2}_{(*)}=12+3^2\\\\x^2+(y-3)^2=21\\\\\boxed{k=3}$

7 0

3 years ago

Help?? I could really use it

Lynna [10]

Answer:

16.5 m

Step-by-step explanation:

The triangle shown in the figure is a right triangle, so we can use the Pythagoreas Theorem.

a² + b² = c²

a² + 32² = 36²

a² + 1024 = 1296

a² = 272

a = √272

a = 16.5

Answer: 16.5 m

4 0

2 years ago

How do you spin the spinner 66 times what is the theoretical probability of it landing on on a odd number

MrRa [10]

Answer:

A CIRCLE

Step-by-step explanation:

ITS A BALL AND IT SPINS CAUSE IT HAS NO SIDES TO STOP

4 0

4 years ago

Save me the headache

maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

or

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

5 0

3 years ago

Read 2 more answers