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3 years ago

Simplify -|-28| + (-5)^2 -53 53 -3

Mathematics

1 answer:

Bond [772]3 years ago

5 0

Answer:

-3 hope this helps :D

Step-by-step explanation:

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How much would $200 invested at 6% interest compounded annually be worth after 6 years? Round your answer to the nearest cent.

Neko [114]

$A=P(1+ \frac{r}{n})^{nt}$
A=future amount
P=present amount
r=rate in decimal form
n=number of times per year to be compounded
t=time in years

so
P=200
r=0.06
n=1
t=6

evaluate to find A
$A=200(1+ \frac{0.06}{1})^{(1)(6)}$
$A=200P(1.06)^{6}$
A=283.704
round
A=$283.70

6 0

3 years ago

Pine Street intersects Center Street at a 65 degree angle. If center Street is parallel to First Avenue, which equation is true?

Nimfa-mama [501]

The answer would be x=180-65 this is because vertically opposite angle to 65 would be 65 which means that x and 65 are supplementary.

5 0

4 years ago

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Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: ordinary differential equation ode integration separable variables initial value problem differential integral calculus

7 0

3 years ago

The length of a rectangle is 3yd longer than its width. If the perimeter of the rectangle is 50yd, find it’s length and width

bazaltina [42]

Answer:

3(15)

Step-by-step explanation:

3 0

3 years ago

HELP NOW PLEASE!! answer choices- 370mm^2 370mm^3 270mm^3 170mm^2

Rama09 [41]

Answer: 370

Step-by-step explanation:

8 0

3 years ago

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