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4 years ago

A. Circle

Mathematics

2 answers:

Shtirlitz [24]4 years ago

6 0

Answer:

it is option A bro

follow me if it is helpful

erma4kov [3.2K]4 years ago

5 0

I thin the answer is C tell me if it rong

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Solve: -12 = 5(p – 2).

bixtya [17]

-12=5p-10
5p=-2
p=-2/5

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3 years ago

Polynomial and rational functions can be used to model a wide variety of phenomena of science, technology, and everyday life. Ch

Shalnov [3]

Answer:

L = s^2/(30.25Cd)

Step-by-step explanation:

In accident investigation, the speed of a vehicle can be estimated using a polynomial function that relates speed (s) to the length of skid marks (L). The drag coefficient Cd will depend on the condition of the road surface and tires, but might be expected to be between 0.7 and 0.8.

If the skid marks end in a collision, the length of the marks that might have been made can be estimated using this formula, then that length added to the actual length of marks to estimate the original speed. The speed at the point of collision can be estimated by the damage caused, and/or the movement created.

In the above formula, length is in feet, and speed is in miles per hour.

8 0

3 years ago

Find the area of a right triangle with side lengths 12, 35, and 37. The length of the altitude to the hypotenuse is 11.35

Snowcat [4.5K]

I got 209.975 units squared

3 0

4 years ago

Wendy won 16% of the games she played. If she won 40 games, how many did she play?

Liula [17]

Let x = total games played

16% = 0.16

0.16x = 40. Divide each number by 0.16.

x = 250

Wendy played 250 games

7 0

4 years ago

Which trigonometric ratios are correct for triangle ABC? Check all that apply

sergij07 [2.7K]

Consider an angle M with measure m≠90°, in a right triangle.

Let

OPP denote the length of the side opposite to M,
ADJ denote the length of the side adjacent to M, and
HYP denote the hypotenuse.

then:

Sin(M) = OPP/HYP
Cos(M)= ADJ/HYPP
Tan(M)=OPP/ADJ

Back to our problem,

using the Pythagorean we can find the length of AB:

$|AB|^2+|AC|^2=|BC|^2\\\\|AB|^2+9^2=18^2\\\\|AB|^2=18^2-9^2\\\\|AB|^2=(18-9)(18+9)=9 \cdot 27=9 \cdot 9 \cdot3\\\\|AB|=9 \sqrt{3} 
$

$Sin(C)= \frac{OPP}{HYP}=\frac{9 \sqrt{3} }{18}= \frac{ \sqrt{3} }{2}$
$Cos(B)= \frac{ADJ}{HYP}= \frac{9 \sqrt{3}}{18}= \frac{ \sqrt{3}}{2}$
$Tan(C)= \frac{OPP}{ADJ}= \frac{9 \sqrt{3} }{9}= \sqrt{3}$
$Sin(B)= \frac{OPP}{HYP}= \frac{9}{18}= \frac{1}{2}$
$Tan(B)= \frac{OPP}{ADJ}= \frac{9}{9 \sqrt{3}}= \frac{1}{ \sqrt{3} }= \frac{ \sqrt{3}}{3}$

Answer: 1, 3, 4

5 0

3 years ago