All categories

3 years ago

Please Please, Please Help!!!! The graph shows the solution for which inequalities?

Mathematics

2 answers:

qwelly [4]3 years ago

5 0

The third one is the answer.

ziro4ka [17]3 years ago

4 0

Hello,

The correct answer is option four:

y ≤ 3x-2 and y ≥ 1/2x+3

Faith xoxo

You might be interested in

If f(x)=3x^2+2×=1 find f(x+2)

harkovskaia [24]

Answer:

f(x+2) = 3(x+2)^2 + 2(x+2 + 1

Step-by-step explanation:

you need to replace the x with x+2

3 0

3 years ago

Please help asap! Will give brainliest! Please answer correctly!<br> No guessing!

Greeley [361]

Answer:

Option B

Step-by-step explanation:

I solved in the picture

Hope this helps ^-^

5 0

3 years ago

Read 2 more answers

The house numbers on the north side of Flynn Street are even. In one block, the house numbers begin 1022, 1032, 1042, and 1052.

never [62]

Answer:

6 houses away

Step-by-step explanation:

You can write down the house numbers and count them:

1032, 1042, 1052, 1062, 1072, 1082

1082 is the 6th house from where Taylor is.

___

You can also recognize that (1082 -1022)/10 = 60/10 = 6 is the number of houses away from the house at 1022.

4 0

3 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .