Can anyone solve this problem please help me

Julia can finish a 20-mile bike ride in 1.2 hours. Katie can finish the same bike ride in 1.6 hours. To the nearest tenth of a m

Brums [2.3K]

Answer:

Step-by-step explanation:

julia : 20 miles in 1.2 hrs = (20/1.2) = 16.67 miles per hr

katie : 20 miles in 1.6 hrs = (20 / 1.6) = 12.5 miles per hr

how much faster does julia ride then katie...

16.67 - 12.5 = 4.17 rounds to 4.2 miles per hr faster <==

4 0

3 years ago

Oliwia made a scaled copy of the following triangle. She used a scale factor greater than 111. What could be the height of the s

GaryK [48]

Answer:

c,d,e

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The functions below are transformations of the parent function f(x) = 3x. Match each function to it's correct transformation.

topjm [15]

Answer:

there all vertical

Step-by-step explanation:

3 0

3 years ago

A tennis star was hired at a convention. She was paid $3000 plus 2% of all admission fees collected at the door. The total amoun

Umnica [9.8K]

3440-3000=440
440=2%

2% x 50 = 100%
440 x 50 = 22000

5 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