Ask question

Ask question

All categories

erastovalidia [21]

3 years ago

13

Solve for x. 2x+20,2x-4

1 answer:

Ket [755]3 years ago

3 0

Are You Saying 2x + 20 = 2x - 4 ? If So There Is No Solution.

You might be interested in

What is the solution to the system of equation below Y equals -1/3 X +6 and X equals -6

Evgen [1.6K]

Answer:

y = 8

Step-by-step explanation:

y= $\frac{1}{3} x$ + 6

$x$ = -6

y=( $\frac{1}{3}$ x 6) + 6

y= 2+6 = 8

6 0

2 years ago

A washer and dryer cost a total of $1152. The cost of the washer is two times the cost of the dryer. Find the cost of each item.

ASHA 777 [7]

Answer: 576 for both

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

SELECT ALL: The point (0, 10) is a solution to the following system of inequalities. Given that information, what could be the s

r-ruslan [8.4K]

63is the number Sorry I don’t know

8 0

3 years ago

Evaluate b2 + 4 · 6<br><br> the 2 is a exponent

Fantom [35]

<h3>Answer:</h3>

b^2+(4•6)
b^2+24

7 0

2 years ago

Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

4 0

3 years ago