All categories

3 years ago

Solve for b in the literal equation a b = c.

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

7 0

Answer:

b = $\frac{c}{a}\\$

Step-by-step explanation:

divide both sides by a

bring the a over to the other side and put it under the c as the opposite of times is divide

You might be interested in

HELP!!! I WILL MARK BRAINLEST!

agasfer [191]

Answer:

B: (16,32)

Step-by-step explanation:

Please give brainlest

4 0

3 years ago

Read 2 more answers

How to solve 2x-1/2=3-x

ValentinkaMS [17]

To solve for x:
Move all the terms containing "x" to the left side of the equation. Do this by adding x to both sides.

2x+x=3x. The new equation is:
3x-1/2=3

Now, move all terms not containing "x" to the right side of the equation. Do this by adding 1/2 to both sides.
3x=3+1/2 The new equation is:
3x=3 1/2, or 3.5

The final step is to isolate x. To do so, divide each side by 3.
3x/3 = 3 1/2 /3 The new equation is:
x=1 1/6

5 0

3 years ago

Algebra find the unknown digit to make each statement true 2.48>2.4 1>2.463

valentina_108 [34]

We have 2.48>2.4 1>2.463. If we insert a '7' between that 4 and 1, we get:

2.48 > 2.471 >2.463, and this is true.

8 0

3 years ago

9.4 The heights of a random sample of 50 college stu- dents showed a mean of 174.5 centimeters and a stan- dard deviation of 6.9

gladu [14]

Answer: a) (176.76,172.24), b) 0.976.

Step-by-step explanation:

Since we have given that

Mean height = 174.5 cm

Standard deviation = 6.9 cm

n = 50

we need to find the 98% confidence interval.

So, z = 2.326

(a) Construct a 98% confidence interval for the mean height of all college students.

$x\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=(174.5\pm 2.326\times \dfrac{6.9}{\sqrt{50}})\\\\=(174.5+2.26,174.5-2.26)\\\\=(176.76,172.24)$

(b) What can we assert with 98% confidence about the possible size of our error if we estimate the mean height of all college students to be 174.5 centime- ters?

Error would be

$\dfrac{\sigma}{\sqrt{n}}\\\\=\dfrac{6.9}{\sqrt{50}}\\\\=0.976$

Hence, a) (176.76,172.24), b) 0.976.

8 0

3 years ago

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago