Ask question

Ask question

All categories

3 years ago

10

Solve for k 7k+2m = kr + 4m + 3

1 answer:

professor190 [17]3 years ago

6 0

Answer:

$\large\boxed{k=\dfrac{2m+3}{7-r}\ \text{for}\ r\neq7}$

Step-by-step explanation:

$7k+2m=kr+4m+3\qquad\text{subtract}\ 2m\ \text{from both sides}\\\\7k+2m-2m=kr+4m-2m+3\\\\7k=kr+2m+3\qquad\text{subtract}\ kr\ \text{from both sides}\\\\7k-kr=kr-kr+2m+3\\\\7k-kr=2m+3\qquad\text{distribute}\\\\k(7-r)=2m+3\qquad\text{divide both sides by}\ (7-r)\neq0\\\\k=\dfrac{2m+3}{7-r}$

You might be interested in

33°<br> (3m + 3)<br> ent OR Vertical_<br> Angle Relationship: Complementary OR Supplementary

Reptile [31]

Answer:

complementary

Step-by-step explanation:

because the angles add up to 90°

3 0

3 years ago

Given: y = 3x - 4. what is the x-intercept? (0, -4) (0, 4/3) (4/3, 0)

ankoles [38]

X intercept is 4/3 because you plug zeros in for y to find the y intercept

4 0

3 years ago

Which value of x would make TV || QS ?<br><br> A. 3<br> B. 8<br> C. 10<br> D. 11

PSYCHO15rus [73]

I hope this helps you

6 0

3 years ago

Read 2 more answers

Test worth 50 points; test score of 78%?

nataly862011 [7]

The person would have scored 39 out of 50. Hope this helps

7 0

3 years ago

(1+cos2x)/(1-cos2x) = cot^2x

sesenic [268]

We will turn the left side into the right side.

$\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x$

Use the identity:

$\cos 2x = \cos^2 x - \sin^2 x$

$\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x$

$\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x$

Now use the identity

$\sin^2 x + \cos^2 x = 1$ solved for sin^2 x and for cos^2 x.

$\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x$

$\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x$

$\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x$

$\cot^2 x = \cot^2 x$

8 0

3 years ago