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4 years ago

10

James overdraws his checking account. If he currently has a balance of -$45.90 and he deposits a check for $65, what is his bala

nce?

1 answer:

11Alexandr11 [23.1K]4 years ago

6 0

19.1 is your answer

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What is the simplified expression in standard form

Schach [20]

Answer:

$\boxed{\sf - 2 {p}^{2} - 11p - 35}$

Step-by-step explanation:

$\sf \implies - 2 {(p + 4)}^{2} - 3 + 5p \\ \\ \sf \implies - 2( {p}^{2} + 2(p)(4) + {4}^{2} ) - 3 + 5p \\ \\ \sf \implies - 2( {p}^{2} + 8p + 16) - 3 + 5p \\ \\ \sf \implies (( - 2) \times {p}^{2} ) + (( - 2) \times 8p) + ( ( - 2) \times 16) - 3 + 5p \\ \\ \sf \implies (- 2 {p}^{2} ) + ( - 16p ) + (- 32) - 3 + 5p \\ \\ \sf \implies - 2 {p}^{2} - 16p - 32 - 3 + 5p \\ \\ \sf \implies - 2 {p}^{2} + (- 16p + 5p) + ( - 32 - 3) \\ \\ \sf \implies - 2 {p}^{2} - 11p - 35$

4 0

4 years ago

Can someone help me find the value of X

Anna007 [38]

X = 31.

because angles in a triangle add up to 180

6 0

3 years ago

Read 2 more answers

Please answer this simple question !!!

Norma-Jean [14]

First, coordinate Q is (-5,2) and Q' is (6,2). First, we know that this is a horizontal reflection becuase only the x-values change. This means that the line of reflection will be x=some value. I think the best way to go about this is to find the midpoint of Q and Q' using the formula: M=(x1+x2/2,y1+y2/2)
Q is the first point and Q' is the second point
M=(-5+6/2,2+2/2)
M=(1/2,4/2)
M=(1/2,2)

Since we know the y-values of Q, Q', and the midpoints are all 2, then the line of reflection would be x=1/2

Hope this helps

8 0

3 years ago

there are 18 girls and 22 boys in a class that has a letter the probability that the owner of the letter is a boy

Slav-nsk [51]

18 + 22 = 40 total kids.

22 are boys

Probability the letter is with a boy would be 22/40 = 11/20

4 0

4 years ago

Solve the differential equations dy/dx=((xy^2)+x)/y

arsen [322]

$\displaystyle \dfrac{dy}{dx}=\dfrac{xy^2+x}{y}\\ \dfrac{dy}{dx}=\dfrac{x(y^2+1)}{y}\\ \dfrac{y}{y^2+1}\, dy=x\, dx\\ \int \dfrac{y}{y^2+1}\, dy=\int x\, dx\\ \dfrac{\ln (y^2+1)}{2}=\dfrac{x^2}{2}+C\\ \ln(y^2+1)=x^2+C\\ y^2+1=e^{x^2+C}\\ y^2=e^{x^2+C}-1\\ y=\sqrt{e^{x^2+C}-1} \vee y=-\sqrt{e^{x^2+C}-1}\\ \boxed{y=\sqrt{Ce^{x^2}-1} \vee y=-\sqrt{Ce^{x^2}-1}}\\$

7 0

4 years ago