The radius of a circle is 3 cm what is the circumference

The director of a community recreation center conducted a six-week study to examine the effects of four types of exercise—streng

Olin [163]

Answer: The treatments in the study are strength training, flexibility training, aerobics, and jogging.

Step-by-step explanation:

A correct description of a feature of the study is that the treatments in the study are strength training, flexibility training, aerobics, and jogging.

It should be noted that the study has no replication. Also, the experimental units are the 40 members that participated.

Therefore, the correct option is D.

4 0

2 years ago

Given the equation of the circle, identify the coordinates of the center. (x+2)2+(y-6)2=49

sveta [45]

Answer:

(-2, 6)

Step-by-step explanation:

Equation of a circle is:

(x − h)² + (y − k)² = r²

where (h, k) is the center and r is the radius.

In this case, the center is (-2, 6).

3 0

2 years ago

3u=48-5u solve for u simplify your answer as much as possible

9966 [12]

U=6 u add 5u and u get 8u on the left side then u divide by 8 and u get 6 so U=6

6 0

2 years ago

In ΔWXY, the measure of ∠Y=90°, the measure of ∠X=74°, and YW = 21 feet. Find the length of XY to the nearest tenth of a foot.

kherson [118]

Answer:

6

this is the right answer

4 0

2 years ago

Can you please answer the question?

Roman55 [17]

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago