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adell [148]
3 years ago
6

Alyssa used a can of paint that was 5/6 full to touch paint her house. When she was done, the bucket was 11/30 full. What fracti

on of the can did she use?
Mathematics
1 answer:
cupoosta [38]3 years ago
3 0

Answer:

7 / 15

Step-by-step explanation:

Fraction before painting = 5/6

Fraction left after painting = 11/30

Fracrion used = difference of the fraction before painting and fraction left after painting

Fraction used = 5/6 - 11/30

Lowest common factor of 6 and 30 = 30

5/6 - 11/30 = (25 - 11) / 30 = 14/ 30

Fraction used = 14/30 = 7/15

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2a(5-b)=3b+7 solve for b<br> Step by step
Sidana [21]
<span>2a(5-b)=3b+7

<em>[open bracket]</em>
10a - 2ab = 3b + 7

<em>[solve for b, so we need to move all terms with b to the left]</em>
<em>[-3b on both sides]</em>
10a - 2ab - 3b = 3b + 7 - 3b
10a - 2ab - 3b = 7

<em>[move all those without b to the right]</em>
<em>[-10a on both side]</em>
10a - 2ab- 3b - 10a = 7 - 10a
-2ab - 3b = 7 - 10a

<em>[divide by -1 through to change b to be positive]</em>
2ab + 3b = 10a - 7

<em>[take b out as the common factor]</em>
</span>b(2a + 3) = 10a - 7<span>

<em>[divide by (2a+3) through]</em>
b = (10a -7)/(2a+3)


</span>
4 0
3 years ago
Lara reads 10 chapters in 4 hours. At this pace, how long will it take her to read 25 chapters? Enter your answer in the box.
Oduvanchick [21]
Lara can read 10 chapters in 4 hours.

At this rate, she can read half that many chapters in half the time.
So she reads 5 chapters in 2 hours.

So Lara reads 10 chapters in 4 hours,
+ 10 chapters in 4 more hours,
+ 5 chapters in 2 more hours.

That's 25 chapters in total, and 10 hours in total, ya? :)


8 0
3 years ago
Read 2 more answers
0.59 O 2/3<br><br> A. =<br> B. &gt;<br> C. &lt;<br> D. none of the above
egoroff_w [7]
0.59 and 2/3

0.59
0.66

0.59 < 2/3

2/3 is greater than 0.59.

Answer :
6 0
3 years ago
20 points
CaHeK987 [17]

Answer:

!!!

Step-by-step explanation:

7 0
2 years ago
Evaluate the line integral, where c is the given curve. c y3 ds,
postnew [5]
Parameterizing \mathcal C by

\mathbf r(t)=(t^3,t)

with 0\le t\le2, we have

\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt=\sqrt{(3t^2)^2+1^2}\,\mathrm dt=\sqrt{9t^4+1}\,\mathrm dt

So

\displaystyle\int_{\mathcal C}y^3\,\mathrm ds=\int_{t=0}^{t=2}t^3\sqrt{9t^4+1}\,\mathrm dt
=\displaystyle\frac1{36}\int_0^236t^3\sqrt{9t^4+1}\,\mathrm dt
=\displaystyle\frac1{36}\int_{u=1}^{u=145}\sqrt u\,\mathrm du
=\dfrac{145^{3/2}-1}{54}
5 0
3 years ago
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