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Viktor [21]
3 years ago
14

Divide. (3a3 + 6a2) ÷ 3a  how do I do this?

Mathematics
1 answer:
yuradex [85]3 years ago
6 0
\frac{ 3a^3 + 6a^2 }{3a}=\frac{3a ( a^2 + 2a ) }{3a}= a^2 + 2a =a(a+2)


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I really need help please
Mkey [24]
So.. let's say ∡AGB and ∡BGC are say "a" units wide, and ∡CGD and ∡DGE are "b" units wide

notice the picture below

all angles added up together, will make up just a flat line, or 180°

now, notice in the picture, the ∡BGD is really just " a + b " wide, notice the green angle in the picture of ∡BGD, well, we know what a + b is

7 0
3 years ago
Convert.<br> 1,000 grams =<br> kilograms
katrin2010 [14]

Answer:

1000 g = 1 kg

Step-by-step explanation:

1000 GRAMS IS EQUAL TO 1 KILOGRAM

4 0
3 years ago
Read 2 more answers
Compare 2/5 and 1/10. what is the denominator
omeli [17]
1/5 five being the denominator, because in math you must always simplify
7 0
3 years ago
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PLEASE HELP!! I WILL MARK BRAINLIST
marishachu [46]

Answer:

m∠5 = 44°; m∠7 = 44°

Step-by-step explanation:

Angle 4 and 1 are supplementary (they make up a line) and their sum is equal to 180 degrees.

Subtracting the measure of angle 4 from 180 degrees gives the measure of angle 1. (180 - 136 = 44).

So Angle 1's measure is 44 degrees.

According to the Corresponding Angle Postulate, Angle 1 and Angle 5 are congruent. Therefore, m∠5 = 44°

According to the Vertical Angles Postulate (if two angles are vertical, they are congruent), ∠5 ≅ ∠7, meaning that m∠5 = m∠7.

So m∠7 = 44°

6 0
2 years ago
Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t &gt; 0.
Margarita [4]

Answer:

y = 3sin2t/2 - 3cos2t/4t + C/t

Step-by-step explanation:

The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

= tsin2t/2 - cos2t/4 ..... 2

Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0
3 years ago
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