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NemiM [27]
2 years ago
9

If anybody willing to help me with my geometry work that’s done by 6pm please contact me by my number 9014132287

Mathematics
1 answer:
My name is Ann [436]2 years ago
4 0
Do not listen to those bots
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Write an equation of the line that passes through point p and is perpendicular to the line with the given equation
goldenfox [79]
Do you have the specific point??

Remember, if 2 lines are perpendicular, their slopes are opposite reciprocals. So if one line has a slope of 4, the other line should have a slope of -1/4.

Hopefully your equation is in y=mx+b form. If so,

make sure you know slope (m) and the y-intercept. After this is done, plug in the points from p for y and x, and make sure to turn m into -1/m. Solve for b, and your new equation should be y=(new slope)x+(new y-intercept)
5 0
3 years ago
Find the equation of the line which passes through the point (−2,−5) and is parallel to the line −3x+5y=−4.
leonid [27]
-3x+5y=-4
5y=3x-4
Y=3/5x-4/5


When it parallel to the line y=3/5x-4/5

It can be y=-3/5x+b

-5=6/5+b

B=-5-6/5=-25/5-6/5=-31/5

Y=-3/5x-31/5

3/5x+y=-31/5

Multiply 5

3x+5y=-31
4 0
3 years ago
Read 2 more answers
YOU WILL GET BRAINLIEST!! 20 POINTS
Mariana [72]

Answer:

Step-by-step explanation:

6⁻︎¹(4⁻︎²)

1/6x1/4 2

1/6 x 1/16

1/96

4 0
2 years ago
Read 2 more answers
Calculate the circumference of the circle. Use 3.14 for pi. I WILL MARK BRAINLIEST
Harrizon [31]

Answer:

The answer would be B

Step-by-step explanation:

C=2(pi)r

C=25.12

8 0
2 years ago
Read 2 more answers
Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2
Finger [1]

Answer:

\frac{y}{x^2}=\sin x+\pi

Step-by-step explanation:

Consider linear differential equation \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)

It's solution is of form y\,I.F=\int I.F\,q(x)\,dx where I.F is integrating factor given by I.F=e^{\int p(x)\,dx}.

Given: \frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x

We can write this equation as \frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x

On comparing this equation with \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x), we get p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x

I.F = e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}      { formula used: \ln a^b=b\ln a }

we get solution as follows:

\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C

{ formula used: \int \cos x\,dx=\sin x }

Applying condition:y(\pi)=\pi^2

\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C

So, we get solution as :

\frac{y}{x^2}=\sin x+\pi

4 0
3 years ago
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