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kirill115 [55]
3 years ago
13

PLS HELP ASAP!!! <3

Mathematics
1 answer:
satela [25.4K]3 years ago
7 0
The answer would be C. A is the same angle and Lines AE=AB and AC=AD
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The parking garage at the airport has 4750 empty parking spaces and 250 full parking spaces. What percent of the spaces in the g
KiRa [710]

Answer:

I'd say around 95% or 94.737%

Step-by-step explanation:

4750 - 250 = 4500

4750/250 = 1/19

1/19 = 0.052 or close to 5%

4750 x 94.737% = 4500.0075

I hope this helps!

3 0
2 years ago
The perimeter of a chalkboard is 16 feet. The area is 15 square feet. What are the dimensions of the board
Daniel [21]

Answer:

5 by 3

Explanation:

15 = 5 x 3

16 = 5 + 5 + 3 + 3

4 0
3 years ago
12×2+7x=<br>I need help understanding why the x is there
posledela
12*2+7(x) you need to solve for x
8 0
2 years ago
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
What is the value of the expression with g=1.5? 5g^2
worty [1.4K]

Answer:

15

Step-by-step explanation:

5 0
2 years ago
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