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2 years ago

8

Please help me with this

1 answer:

Naya [18.7K]2 years ago

7 0

Answer:

the geratest distance is

10m at the time 20 sec

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Charlie is at an amusement park that has 18 different rides. Each ride costs $1.75. The function C(r) = 1.75r represents the tot

sleet_krkn [62]

Domain= [0,18], Range = [0,31.50]
If Charlie will only go to a single ride one time the possible rides he could go on would be 0 rides to all 18 rides for a domain of [0,18]. The output of the maximum and minimum x values would give us 0 dollars for riding 0 rides and 31.50 for riding all 18 (1.75 *0 = 0) and (1.75*18 = 31.50) so the range is [0, 31.50]

4 0

3 years ago

Multiple choice please answer please

juin [17]

Answer:

D

Step-by-step explanation:

Using the sine ratio in the right triangle

sin B = $\frac{opposite}{hypotenuse}$ = $\frac{AC}{AB}$ = $\frac{2}{9}$ , thus

B = $sin^{-1}$ ( $\frac{2}{9}$ ) ≈ 12.84° → D

4 0

3 years ago

Bye thanks for the help got fam issues

nika2105 [10]

Answer: okay, bye

Step-by-step explanation:

8 0

3 years ago

If A(3,4) and B(10,4) are the end points of a diameter of circle "O", what is the area of the circle O

ikadub [295]

Answer:

$\large\boxed{\dfrac{49\pi}{4}}$

Step-by-step explanation:

The formula of an area of a circle:

$A=\pi r^2$

<em>r</em> - radius

We have the endpoints of a diameter of a circle.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the given points A(3, 4) and B(10, 4):

$d=\sqrt{(10-3)^2+(4-4)^2}=\sqrt{7^2+0^2}=\sqrt{7^2}=7$

Therefore we have the length of the diameter: <em>d = 7</em>

We know: <em>d = 2r.</em> Therefore

$r=\dfrac{7}{2}$

Substitute to the formula of an area:

$A=\pi\left(\dfrac{7}{2}\right)^2=\dfrac{49\pi}{4}$

<em>There is no such answer in the answers to choose.</em>

<em>Apparently you have some error in the coordinates of the endpoints.</em>

8 0

3 years ago

What is 2(5)+3(4)+5^2? (The ^ means the exponent)

castortr0y [4]

I believe the answer is 47.

6 0

3 years ago