Use AABC to find the value of sin B.

Houses can be either domed or octagonal. True False

QveST [7]

Answer:

true

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Solve for b −3b+2.5=4 Two step equations with fractions and decimals

DIA [1.3K]

−3b+2.5=4

Subtract 2.5 from both sides.

−3b+2.5−2.5=4−2.5

−3b=1.5

Divide both sides by -3.

−3b/−3 = 1.5/−3

b=−0.5

3 0

3 years ago

Read 2 more answers

What is 4 1/2 pint to quart?

Ivanshal [37]

There are two pints every quarter.

So let's convert pints to quarter

$\frac{4\frac{1}{2} }{2} =\frac{\frac{9}{2} }{2} = \frac{9}{4}$

There 4 1/2 pint is to 9/4 quarter.

Hope that helps!

p.s. look at my diagram to help you remember

5 0

2 years ago

How many pounds of raisins were in her supply when she started?

sp2606 [1]

Answer:

i think it's 15 because if you use 5 and you have to do it by 3 it should be 15

8 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