Ask question

Ask question

All categories

2 years ago

13

Selene's checking account balance was -$33, and then she withdrew $42 more. What is her checking account balance after the withd

rawal?

2 answers:

Rainbow [258]2 years ago

7 0

Answer:

-75

Step-by-step explanation:

WIthdrew = Negative

-33 - 42 = -75

xxMikexx [17]2 years ago

3 0

Answer: $-9

Step-by-step explanation:

$33-$42 = $-9

I believe that is the answer

You might be interested in

Which of the following steps would you perform to the system of equations below so that the equations have opposite y-coefficien

denis23 [38]

I think B not 100%shure ,you could prob go ask this on yahoo and get a bit more help ;) sorry I couldent help more

3 0

3 years ago

Read 2 more answers

zhenek [66]

- Trapezoid

The measures of three angles of a quadrilateral are 70, 125, and 89 degrees. Find the measure of the fourth one.

》The measure of the fourth one is 36°.

<h3>HOPE ITS HELP</h3>

4 0

2 years ago

Please help with A,B for 10 points.

denis-greek [22]

Answer:

y= (C -Ax)/B

x= (C -By)/A

Step-by-step explanation:

Ax+ By = C

solve for y

Subtract AX from each side

Ax+ By - Ax = C-Ax

By = C -Ax

Divide by B

By /B= (C -Ax)/B

y= (C -Ax)/B

solve for x

Subtract By from each side

Ax+ By - By = C-By

Ax = C -By

Divide by A

Ax/A= (C -By)/A

x= (C -By)/A

4 0

2 years ago

Read 2 more answers

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

seropon [69]

Answer:

$h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

$g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b$

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

$h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

3) To test it, just integrate:

$\int \frac{3r^{2}}{2\sqrt{r^3+5}}dr=\sqrt{r^{3}+5}+C$

5 0

2 years ago

What are the zeros of the function defined by 2x + 11x- 6?

asambeis [7]

Note --> As you're writing 'zeroes' , I assume there is a typing mistake and you've given a Quadratic equation in the question

$f(x) = 2 {x}^{2} + 11x - 6 \\ \\ f(x) = (2x - 1)(x + 6) = 0$

Hence , the roots are 1/2 and -6
The graph attached confirms the same ,

Hope it helps you :)

6 0

3 years ago