All categories

3 years ago

The length of a rectangle is 5 yd more than twice the width x. The area is 900 yd?.

Mathematics

1 answer:

givi [52]3 years ago

8 0

A = l * w. w = x; l = 5 + 2x
(5+2x)x=900

You might be interested in

What two number multiplies to get 15 and adds to get 6

OverLord2011 [107]

Asking the Math Gods...

The factors of 15 Answer : 1,3,5,15

so 3 and 5 are the only that would get 15

therefore this problem isnt correctly written.. You can't get 6 by adding any of the factored numbers

5 0

3 years ago

A store bought 300 toys at a cost of $24 each. The store sold all the toys at a percent markup of . Find the total selling pric

BlackZzzverrR [31]

Answer:

Step-by-step explanation:

You will have to write out the following:

25% of 24= ?

Then go ....turn 25 into a decimal

.25 x 24

x.25

——— ——) Which equals 6

Follow these steps on any one and as long as ur multiplication is right the answer will be right

3 0

3 years ago

What is 50 percent of 18?

Effectus [21]

Steps:
18*0.5=9

Answer:

9

3 0

2 years ago

Read 2 more answers

I really need help with this

kupik [55]

Answer:

Oh Stop!!!! This is sooooooooooooo easy! If you would just pay attention in school and not spend all your time trying to buy the new ipad pro maybe you would not have to go to Branily to get a simple answer! Do you think your teacher would like you cheating like this???? I dont think so. So get of of Branily and start studying!!!!

Step-by-step explanation:

4 0

2 years ago

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

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

The marked are the trigonometric identities

3 0

1 year ago