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Anton [14]
3 years ago
6

Solve for W. 4w-4+2(5w+8)=-2(w+3) Simplify your answer as much as possible.

Mathematics
2 answers:
gayaneshka [121]3 years ago
8 0

Answer:

w = -\frac{9}{8}

Step-by-step explanation:

The question is  4w-4+2(5w+8)=-2(w+3)

We can first use distributive property to simplify, the distributive property is  a(b+c)=ab+ac

Thus we have:

4w-4+2(5w+8)=-2(w+3)\\4w-4+10w+16=-2w-6

<em>Now we combine like terms and simplify to get the final answer for w</em><em>.</em>

4w-4+10w+16=-2w-6\\14w+12=-2w-6\\14w+2w=-6-12\\16w=-18\\w=\frac{-18}{16}\\w=-\frac{9}{8}

Thus, w = -\frac{9}{8}

murzikaleks [220]3 years ago
7 0

Answer:

1.125

Step-by-step explanation:

we first open the brackets by multiplying every term inside the brackets by 2

We get the following equation: 4w-4+10w+16=-2w-6

collecting like terms together we get 4w+10w+2w=-6-16+4

   16w=-18

dividing both sides by 16 we get w=18/16

=9/8

= 1 1/8 or  1.125

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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.

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4 years ago
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