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pentagon [3]
3 years ago
15

HELP! PLEAAAAAAAAASEE

Mathematics
1 answer:
DerKrebs [107]3 years ago
4 0
What's the question?
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The figure shows five polygons: ABCDE, FGHIJ
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It can transform in ABCDE
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G(x) = 4x + 3; Find g(10)
adelina 88 [10]

Answer:

43

Step-by-step explanation:

g(x)=4x+3

plug in 10

g(10)= 4(10)+3

      = 40+3

      = 43

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180 product of a prime factor using indices
zvonat [6]
Start with 180. 
<span>Is 180 divisible by 2? Yes, so write "2" as one of the prime factors, and then work with the quotient, 90. </span>

<span>Is 90 divisible by 2? Yes, so write "2" (again) as another prime factor, then work with the quotient, 45. </span>

<span>Is 45 divisible by 2? No, so try a bigger divisor. </span>
<span>Is 45 divisible by 3? Yes, so write "3" as a prime factor, then work with the quotient, 15 </span>

<span>Is 15 divisible by 3? [Note: no need to revert to "2", because we've already divided out all the 2's] Yes, so write "3" (again) as a prime factor, then work with the quotient, 5. </span>

<span>Is 5 divisible by 3? No, so try a bigger divisor. </span>
Is 5 divisible by 4? No, so try a bigger divisor (actually, we know it can't be divisible by 4 becase it's not divisible by 2)
<span>Is 5 divisible by 5? Yes, so write "5" as a prime factor, then work with the quotient, 1 </span>

<span>Once you end up with a quotient of "1" you're done. </span>

<span>In this case, you should have written down, "2 * 2 * 3 * 3 * 5"</span>
5 0
3 years ago
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25 points<br><br> What type of graph is best to use to compare two sets of data over time?
katrin [286]

Answer:

Bar Graph

Step-by-step explanation:

Bar graphs are used to compare things between groups or tracks. They are the most efficient because the set data is set to change at any time

3 0
3 years ago
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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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