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monitta
3 years ago
13

Lisa has entered a check payment of $213.75 in her checkbook. If the old balance was $794, what is her latest balance? $1,007.75

$581.75 $580.25
Mathematics
2 answers:
Digiron [165]3 years ago
3 0
$580.25
You subtract $213.75 out of the $794.
hodyreva [135]3 years ago
3 0

Answer:

c.580.25

Step-by-step explanation:

plato lives matter

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Write a function<br> Rule for the output<br> Is seven more than<br> the input
Nadusha1986 [10]

Answer:

f(x) = x + 7

or

y=x+7

Step-by-step explanation:

7 0
3 years ago
How does -1.125 + 1.2 = 0.075 instead of 0.925?
kenny6666 [7]

Answer:

.0075 (hope this help; let me know if not)

Step-by-step explanation:

The operation is + and the signs of the numbers are opposite so we have to subtract. The bigger number has positive sign in front if so we do 1.2-1.125=1.200-1.125

I'm going to line up

1.200

-1.125

---------

If you have 200 and you take about 125... 75 should beleft

so

1.200

-1.125

---------

0.075

6 0
3 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
X = y - 3
Leviafan [203]
The answer is A. (1, 4), because when the values are substituted in to the equations, you get 1 = 4 - 3 and 1 + 12 = 13, which are both correct. I hope this helps!
3 0
3 years ago
What is the bi-conditional statement of the conditional statement below? What is the truth value?
ozzi

Answer:

I doesn't know but, I blow your junk

Step-by-step explanation:

Because it could be any of these

6 0
3 years ago
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