1/2a=1/4a+2 Solve for a: Check your solution:

13. Maria earns $8 per hour plus a 5% commission on the price

Pani-rosa [81]

Answer:

Inequality:

120 + 0.05x ≥ 200

Solution:

x ≥ $1,600

Her total weekly sales must be equal to or greater than $1,600

Step-by-step explanation:

Let x represent the weekly sales she must make to reach her goal.

Given;

Pay rate = $8

Weekly total work hours = 15 hours

Commission on sales = 5% = 0.05

Total weekly earnings is;

8×15 + 0.05×x

120 + 0.05x

Minimum Weekly target earnings = $200

So;

120 + 0.05x ≥ 200

Solving the inequality equation;

0.05x ≥ 200 - 120

0.05x ≥ 80

x ≥ 80/0.05

x ≥ 1600

x ≥ $1,600

Her total weekly sales must be equal to or greater than $1,600

6 0

3 years ago

Need help!!! will mark brainliest!!!!!!

irga5000 [103]

Answer:

right 1 unit and down 5 units

Step-by-step explanation:

5 0

3 years ago

Help! This is due today!

bulgar [2K]

Answer:

27π

Step-by-step explanation:

area of circle = πr^2

= π × 6^2

= 36π

36/4

= 9

36 - 9

= 27π units squared

4 0

2 years ago

Carl has 9 1/2 cups of juice for a party. If each party goer gets exactly 3/5 cup of juice each, how many party goers will get t

topjm [15]

Answer: 158

Step-by-step explanation:

Okay so the total is 9 5/10.

Each party goer needs 6/10

So 95/10 / 6/0=. 95/10x10/6

=950/6 = 158.

7 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