I WILL GIVE YOU BRAINLIEST IF YOU GET IT RIGHT! YOU WILL ALSO GET 10 POINTS IF YOU ATTEMPT!

Mathematics

1 answer:

mylen [45]3 years ago

4 0

Answer:

B. 120

Step-by-step explanation:

<u>so his average speed is 20 mph in heavy traffic, and 50 mph in light traffic</u>

in 1 hour of heavy traffic, he would have gone 20 miles because if his average speed is 20 mph (miles per hour) in heavy traffic, then in 1 hour he would have gone 20 miles

20(1) = 20

in 2 hours of light traffic, he would have gone 100 miles because if his average speed is 50 mph (miles per hour) in light traffic, then in 2 he would have gone 100 miles

50(2) = 100

in total, if you add the 20 and 100, you find out how far abdul traveled:

20 + 100 = 120

so in 1 hour of heavy traffic and 2 hours in light traffic, abdul would have traveled 120 miles

You might be interested in

Please answer due today (one question) Algebra and WILL GIVE BRAINLEST!!!

chubhunter [2.5K]

180(6 - 2) = 720

4x + 180 = 720

4x = 630

x = 157.5

S = 720

X = 157.5

5 0

3 years ago

Write the equation of a line with a slope of -2 and a y-intercept of 5:

saveliy_v [14]

Answer: y = -2x +5 , so C (you probably just forgot to put the x)

Explanation: y = mx+b m represents the slope, b represents the y intercept

6 0

3 years ago

Read 2 more answers

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities