What is the written form of the number 62?

AnnZ [28]

Answer:

x = 5/2 y = 3/2

Step-by-step explanation:

You need to write what the question is besides the information in the problem. I assume the question is to find the values of x and y.

Going on that premise, Let equation (1) be -5x + 3y = -8 and

equation (2) be 3x - 7y = -3

Lets multiply (1) by 3 and (2) by 5 and we get equations (4) and (5)

(4) -15x + 9y = -24 and (5) 15x - 35y = -15 Now add (4) and (5)

-15x + 9y = -24

15x - 35y = -15

-26y = - 39

y = 3/2 Now substitute y = 3/2 in equation (1). -5x + 3)(3/2) = -8

-5x + 9/2 = -8

-10x + 9 = -16

-10x = -25

x = -25/-10 = 5/2

x = 5/2 y = 3/2

Check: Substitute your answers into equation (2) and see if they work.

3(5/2) -7(3/2) = 15/2 - 21/2 = -6/2 = -3 Hooray! We have the correct values for x and y that makes each equation true

8 0

3 years ago

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:

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

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

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

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