Find the area of the square ABCD. Need green and grey box! Please help me! Sorry for the bad quality lol

1

svetoff [14.1K]

Answer:

1/2 ornament in an hour

Step-by-step explanation:

Given

$Orna\ ments = 2$

$Time = 4$

Required

Determine the ornaments per hour

This question implies that we calculate the unit rate.

$Unit\ Rate = \frac{Orna\ ments}{Time}$

Substitute values for Time and number of ornaments

$Unit\ Rate = \frac{2}{4}$

$Unit\ Rate = \frac{1}{2}$

This implies that Constance can decorate 1/2 ornaments in an hour

3 0

2 years ago

PLS Write a real world problem for the inequality 8-2x≤5.

Hatshy [7]

Step-by-step explanation:

8-2x< 5
-2x<5-8
x< -3/-2
x<1.5

hope it helps.

3 0

2 years ago

First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x

e-lub [12.9K]

Answer:

$(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

$\int x ln(5+x)dx$

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

$\int (U-5) ln(U)dU$

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

$\int (pq')=pq-\int qp'$

so we must define p, q, p' and q':

p=ln U

$p'=\frac{1}{U}dU$

$q=\frac{U^{2}}{2}-5U$

q'=U-5

and now we plug these into the formula:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU$

Which simplifies to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU$

Which solves to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C$

so we can substitute U back, so we get:

$\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C$

and now we can simplify:

$\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

notice how all the constants were combined into one big constant C.

7 0

3 years ago

In a group of 60 students, 12 students take algebra one, 18 students take algebra 2 and 8 students take both subjects how many s

hoa [83]

60-12-18-2-8=20
=48-18-2-8
=30-2-8
=28-8
=20

So, 20 students don't take either of these subjects.

5 0

3 years ago

I did the math and ended up with 3,033 but that is waaaay over his savings please help

rewona [7]

Answer: Dave Will have $881 left to put into his savings.

Step-by-step explanation

So Lets look at this, He has $2000 to spend, he buys 3 tables for $58 each, 12 chairs for $65 each, and a watch for $165.

(2000-(3*58)-(12*65)-165)=881

8 0

3 years ago

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