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Komok [63]
2 years ago
8

HELP PLEASE THIS IS DUE IN 5 MINUTES

Mathematics
1 answer:
Reika [66]2 years ago
4 0

Answer:

Part A:

Since Jasmine has more cousins rhan Brenda, who has 4 cousins, then the amount of cousins that Jasmine has (j) is greater than 4 so j > 4

Part B:

We have a number line and we draw an arrow starting from 4 and going to the right.

Hope this helped!

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Virty [35]

Answer:

\int\limits^1_0 \int^1_0{z} \, dA

Step-by-step explanation:

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\int\limits^1_0 {9y-4x^2y-\frac{5y^3}{3}|_0 ^1 } \, dx

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You finish the leftover.

7 0
3 years ago
Please help I only have a few minutes left
GalinKa [24]

Answer:

4 because that's where the lines intercept

6 0
3 years ago
(b) dy/dx = (x - y+ 1)^2
Elanso [62]

Substitute v(x)=x-y(x)+1, so that

\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}

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1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx

On the left, we can split into partial fractions:

\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx

Integrating both sides gives

\dfrac{\ln|1-v|+\ln|1+v|}2=x+C

\dfrac12\ln|1-v^2|=x+C

1-v^2=e^{2x+C}

v=\pm\sqrt{1-Ce^{2x}}

Now solve for y(x):

x-y+1=\pm\sqrt{1-Ce^{2x}}

\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}

3 0
3 years ago
The area of a square photo is 9 square inches how long is each side of the photo
Marta_Voda [28]
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So area of photo = (side)^2 = a^2

But as given area of photo is 9 square inch.

So a^2 = 9

a =  \sqrt{9}
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So side of photo is 3 inch.
3 0
3 years ago
What is the order of the numbers from least to greatest? A = 1.5 x 10^3; B = 1.4 x 10^-1; C = 2 x 10^4; D = 1.4 x 10^-2
12345 [234]
A
D=0.014, B=0.14, A=1500, C=20000
7 0
3 years ago
Read 2 more answers
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