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torisob [31]
4 years ago
9

What is the circumference of this circle

Mathematics
2 answers:
sweet-ann [11.9K]4 years ago
6 0
P=pie
P•d= circumference
3p
Yakvenalex [24]4 years ago
3 0

Answer:

C = 9.42cm

Step-by-step explanation:

C = pi*d

C = 3.14 x 3 = 9.42

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What is 1/3 divided by 2/7 equal to
Contact [7]
1/3 / 2/7
The first thing you do is switch the 2 and the 7 around and change the sign to a times: 
1/3 X 7/2

1 X 7 = 7
3 X 2 = 6

1/3 X 7/2 = 7/6

7/6 (which is an improper fraction) can we written as 1 1/6 (which is a mixed number). 

The answer is 1 1/6

8 0
3 years ago
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Question 4
Harman [31]

Answer:

False

Step-by-step explanation:

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m}

Given a line with m = \frac{3}{4}

Then the slope of the line perpendicular to it is

m_{perpendicular} = - \frac{1}{\frac{3}{4} } = - \frac{4}{3}

4 0
4 years ago
Help please if you dont mind​
Bumek [7]

Answer:

D)

Step-by-step explanation:

For every x, there can be only y.

A, B and C violate that.

8 0
3 years ago
How does the graph of y = 2 · 2x compare to the graph of the parent?
Likurg_2 [28]

This can be also written as y=-2x+2

First of all the - at the beginning = reflection across the x axis.

Then the first 2 is a vertical stretch of 2

the second 2 is a translation of 2 units up.

So

1.ref. across the x axis

2.vert. stretch of 2

3. translation of 2 units up

6 0
3 years ago
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
4 years ago
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