The average distance in miles traveled 1hour if 660miles can be covered in 1.5 hours

Joe's hot dog stand is only open for lunch. of today's sales 26% came from selling plain hot dogs. Since the hot dog stand made

meriva

97 then add and find your multiplying

6 0

2 years ago

1

amm1812

Answer:

30m squared carpet is needed to cover the hallway and the living room.

Step-by-step explanation:

Living room area

6 × 4.2 = 25.2

Hallway area

3.2 × 1.5 = 4.8

Total area

25.2 + 4.8 = 30m squared

5 0

2 years ago

Solving Linear Inequalities can someone help me know how to this please

KATRIN_1 [288]

Answer:

how to what? I don't understand

5 0

2 years ago

Read 2 more answers

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

1 year ago

Solve for x. y=c(x+b) Enter your answer in the box.

Novay_Z [31]

Answer:

Step-by-step explanation:

if you could tell me what you're learning, it could help me.

what you have to try to do i believe is you have to isolate the x. you're solving for x, so you're isolating it. you want to get x alone.

x = y/c (y over c) - b

divide both sides by c

x + b = y/c (subtract b from both sides)

x = y/c - b

please give me thanks if this helped!

5 0

2 years ago

Read 2 more answers