All categories

3 years ago

Solve for w. 6w+18-9w = -12 Simplify your answer as much as possible. W=

Mathematics

2 answers:

marusya05 [52]3 years ago

5 0

Answer:

w = 10

Step-by-step explanation:

6w+18-9w = -12

Combine like terms

-3w + 18 = -12

Subtract 18 from each side

-3w +18- 18 = -12 -18

-3w = -30

Divide each side by -3

-3w/-3 = -30/3

w = 10

Blababa [14]3 years ago

4 0

Answer:

w = 10

Step-by-step explanation:

6 w + 18 - 9 w = - 12

combine like terms

6 w - 9 w + 18 = - 12

- 3 w + 18 = - 12

subtract each side by 18

- 3 w + 18 - 18 = - 12 -18

- 3 w = - 30

Divide each side by - 3

- 3 w / - 3 = - 30 / - 3

w = 10

You might be interested in

Rewrite the fraction using the least common denominator <br><br> 4/9 7/15

denis23 [38]

Answer:

20/45 & 21/45

Step-by-step explanation:

Find a common denominator. What you do to the denominator, you do to the numerator. In this case, the smallest denominator is 45.

(4/9)(5/5) = 20/45

(7/15)(3/3) = 21/45

The two fractions you have is:

20/45 for 4/9

21/45 for 7/15

6 0

4 years ago

Read 2 more answers

Which of the following is(are) the solution(s) to 7|X+2|= -49 ?

ziro4ka [17]

Answer:

(B) No Solution

Step-by-step explanation:

This statement is false for any value of it.

4 0

3 years ago

I NEED SOMEONE TO ANSWER THIS QUESTION

Viktor [21]

Answer:

Step-by-step explanation:

5=20% of x

0.20x=5

divide both sides by 0.20

x=25

8 0

3 years ago

1+1+1+1+1 12312123 times

OverLord2011 [107]

Answer:

$12312123$

5 0

3 years ago

Read 2 more answers

Be sure to answer all parts. List the evaluation points corresponding to the midpoint of each subinterval to three decimal place

gayaneshka [121]

Answer:

The Riemann Sum for $\int\limits^5_4 {x^2+4} \, dx$ with n = 4 using midpoints is about 24.328125.

Step-by-step explanation:

We want to find the Riemann Sum for $\int\limits^5_4 {x^2+4} \, dx$ with n = 4 using midpoints.

The Midpoint Sum uses the midpoints of a sub-interval:

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We know that a = 4, b = 5, n = 4.

Therefore, $\Delta{x}=\frac{5-4}{4}=\frac{1}{4}$

Divide the interval [4, 5] into n = 4 sub-intervals of length $\Delta{x}=\frac{1}{4}$

$\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right]$

Now, we just evaluate the function at the midpoints:

$f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625$

$f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625$

$f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625$

$f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625$

Finally, use the Midpoint Sum formula

$\frac{1}{4}(21.015625+23.140625+25.390625+27.765625)=24.328125$

This is the sketch of the function and the approximating rectangles.

5 0

4 years ago