CAN SOMEONE HELP ME ANSWER THIS

The vertex of the parabola below is at the point (2,4), and the point (3,6) is

lukranit [14]

Answer:

A. y= 2(x - 2)2 + 4

Step-by-step explanation:

vertex having an x value of 2 means the phrase in paren must be x - 2.

This eliminates B and D

vertex having a y value of 4 is of no help as the remaining two equations will result in 4 if the x term is 2

plugging in x = 3 makes y = 6 in equation A, but not in equation C where y = 10

3 0

3 years ago

9. A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

SSSSS [86.1K]

Answer:

Part 4) $r=84\ units$

Part 9) $sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) $sin(\theta)=-\frac{9\sqrt{202}}{202}$

Step-by-step explanation:

Part 4) A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

we know that

The circumference of a circle subtends a central angle of 360 degrees

The circumference is equal to

$C=2\pi r$

using proportion

$\frac{2\pi r}{360^o}=\frac{56\pi}{120^o}$

simplify

$\frac{r}{180^o}=\frac{56}{120^o}$

solve for r

$r=\frac{56}{120^o}(180^o)$

$r=84\ units$

Part 9) Given cos(∅)=-2/3 and ∅ lies in Quadrant III. Find the exact value of sin(∅) in simplified form

Remember the trigonometric identity

$cos^2(\theta)+sin^2(\theta)=1$

we have

$cos(\theta)=-\frac{2}{3}$

substitute the given value

$(-\frac{2}{3})^2+sin^2(\theta)=1$

$\frac{4}{9}+sin^2(\theta)=1$

$sin^2(\theta)=1-\frac{4}{9}$

$sin^2(\theta)=\frac{5}{9}$

square root both sides

$sin(\theta)=\pm\frac{\sqrt{5}}{3}$

we know that

If ∅ lies in Quadrant III

then

The value of sin(∅) is negative

$sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) The terminal side of ∅ passes through the point (11,-9). What is the exact value of sin(∅) in simplified form?

see the attached figure to better understand the problem

In the right triangle ABC of the figure

$sin(\theta)=\frac{BC}{AC}$

Find the length side AC applying the Pythagorean Theorem

$AC^2=AB^2+BC^2$

substitute the given values

$AC^2=11^2+9^2$

$AC^2=202$

$AC=\sqrt{202}\ units$

so

$sin(\theta)=\frac{9}{\sqrt{202}}$

simplify

$sin(\theta)=\frac{9\sqrt{202}}{202}$

Remember that

The point (11,-9) lies in Quadrant IV

then

The value of sin(∅) is negative

therefore

$sin(\theta)=-\frac{9\sqrt{202}}{202}$

5 0

3 years ago

Please help me with this

tankabanditka [31]

I think this is what I found

3 0

3 years ago

Note: A maximum number of 4 points are available for this question. Part A is worth 1 point. Part B is worth 1 point. Part C is

vovikov84 [41]

Answer:

There is a decreasing pattern the farther they are the less likely hit the target.

Step-by-step explanation:

4 0

3 years ago

Write a rule for the linear function in the table above. ( Hint: Use the formula y = mx + b )

Marta_Voda [28]

Y=5x-3

the slope is plus 5
so if 1 = 2
simply subtract 5 from two and you get -3

6 0

3 years ago

CAN SOMEONE HELP ME ANSWER THIS ​

CAN SOMEONE HELP ME ANSWER THIS