Since they’re vertical we know that they’re congruent. We write the equation as 160=10x+50. Subtract 50 to both sides then we get 110=10x so now we divide 10 to both sides and get 11=x.
So your answer is A
Hope I helped :)

7 0

3 years ago

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

8 0

3 years ago

Let f(x) = ............

34kurt

Answer:

So first lets find g(-1)

so we plug in the answers:

-2 * -1 ^2 -4

-2*1-4 = -2-4 =

-6

Now lets solve for f(-2)

-2^2-3

4-3=1

-6+1 = 5

3*5 = 15

Im getting answer of 15 but it shows -15 or something, so I dont know if I got it wrong or if its like a dash then the answer.

3 0

3 years ago