What is the area of the shape below?

erastovalidia [21]

Answer:

Do you mean like on the line it would be x=3 or where you looking for a different type of answer?sorry if this isn't too helpful

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

Matthew made a fruit salad, in which the ratio of blueberries to red grapes is 8:6. Write the ratio of red grapes to blueberries

g100num [7]

Answer:

6:8

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Express the square root of -80 in its simplest terms.

spin [16.1K]

Answer:

C

Step-by-step explanation:

noting that $\sqrt{-1}$ = i

Given

$\sqrt{-80}$

= $\sqrt{16(-1)(5)}$

= $\sqrt{16}$ × $\sqrt{-1}$ × $\sqrt{5}$

= 4i $\sqrt{5}$ → C

4 0

3 years ago

Evaluate 6(y-6) when y=9.

julia-pushkina [17]

Answer:

18

Step-by-step explanation:

Substitute 9 as y. 6(9-6). Subtract: 6(3). Multiply: 18

7 0

3 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

2 years ago