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Ivenika [448]
3 years ago
13

I neeed help with middle school math

Mathematics
2 answers:
Artist 52 [7]3 years ago
6 0
Multiple different forms in the picture attached - for these problems use Math-way is super easy
First is the exact form

dexar [7]3 years ago
5 0

Answer:

22/5*8/3

176/15

yep that's it

I think..

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What is three squared minus two thirds?
Tems11 [23]

three squared minus two thirds

3^2-\frac{2}{3}

three square = 3 x 3 = 9

\begin{gathered} 3^2-\frac{2}{3}=9-\frac{2}{3} \\ 3^2-\frac{2}{3}=\frac{9}{1}-\frac{2}{3} \end{gathered}

Least common factor of 1 and 3 is 3

\begin{gathered} 3^2-\frac{2}{3}=\frac{9}{1}-\frac{2}{3} \\ 3^2-\frac{2}{3}=\frac{27-2}{3} \\ 3^2-\frac{2}{3}=\frac{25}{3} \end{gathered}

Three squared minus two third is twenty five by three.

5 0
11 months ago
Will mark u as BRAINLIEST plz help
ale4655 [162]

Answer:

Im not sure

Step-by-step explanation:

this is kinda hard

4 0
3 years ago
Can u pls help I need a good grade
almond37 [142]

Answer:

15?

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Describe the transformations of the given function:<br><br> f(x)=−14|x−2|+5
statuscvo [17]

Answer:

im busy so i cant right now

Step-by-step explanation:

7 0
2 years ago
Find an explicit solution of the given initial-value problem. (1 + x4) dy + x(1 + 4y2) dx = 0, y(1) = 0
MissTica

Answer:

a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

Step-by-step explanation:

for the equation

(1 + x⁴) dy + x*(1 + 4y²) dx = 0

(1 + x⁴) dy  = - x*(1 + 4y²) dx

[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx

∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx

now to solve each integral

I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁

I₂=  ∫[-x/(1 + x⁴)] dx

for u= x² → du=x*dx

I₂=  ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ =  - tan⁻¹ (x²) +C₂

then

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C

for y(x=1) = 0

1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C

since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for  π*N , we will choose for simplicity N=0 . hen an explicit solution would be

1/2 * 0 = - π/4 + C

C= π/4

therefore

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

5 0
3 years ago
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