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Talja [164]
2 years ago
8

Please help i have a timer

Mathematics
2 answers:
sveta [45]2 years ago
7 0
93.75 pints of 90% & 56.25 of 70%
KIM [24]2 years ago
6 0

Answer: 56.25 of 70% and 93.75pints of 90%

Step-by-step explanation: First find the ratios. One juice is 70% and the other is 95%. Your goal is 85%. This is 15% from 70 and 10% from 95. This means you need (15)/(15+10) of 95, which is 3/5 of all the juice to be 95. This means you will need more of the 95% juice, which makes sense because 85 is closer to 95. So 3:5 and total 150=56.25 of 70% and 93.75pints of 90%

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Josh will take 10 tests in 5 weeks at school. how many tests will he have taken in 7 weeks
Virty [35]

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14 tests in 7 weeks

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4 0
2 years ago
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The number of boys varies directly as the number of girls and inversely as the number of teachers. When there are 24 boys and 16
lidiya [134]
SO from what I did was...

Since 24 x 3 = 72 , I did the same to 16. 

16 x 3 = 48 

But because instead of their being 3 teachers there is 1... What I did is just was to divide 48 / 2 which gave me the answer of 24... 

So what I think the answer is... 

There are 72 boys, 24 girls, and 1 teacher... 

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3 0
3 years ago
Which Two equations below are perpendicular?
UkoKoshka [18]

Answer: The answer is the second equation (y = 3x-4 and y = -1/3x - 2)

Step-by-step explanation:

This is because two lines are perpendicular when the slopes of the equations are negative reciprocals from each other.

3 0
2 years ago
The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo
oksano4ka [1.4K]

Answer:

Required center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

Step-by-step explanation:

Given semcircles are,

y=\sqrt{1-x^2}, y=\sqrt{16-x^2} whose radious are 1 and 4 respectively.

To find center of mass, (\bar{x},\bar{y}), let density at any point is \rho and distance from the origin is r be such that,

\rho=\frac{k}{r} where k is a constant.

Mass of the lamina=m=\int\int_{D}\rho dA where A is the total region and D is curves.

then,

m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k

  • Now, x-coordinate of center of mass is \bar{y}=\frac{M_x}{m}. in polar coordinate y=r\sin\theta

\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta

=3k\int_{0}^{\pi}\sin\theta d\theta

=3k\big[-\cos\theta\big]_{0}^{\pi}

=3k\big[-\cos\pi+\cos 0\big]

=6k

Then, \bar{y}=\frac{M_x}{m}=\frac{2}{\pi}

  • y-coordinate of center of mass is \bar{x}=\frac{M_y}{m}. in polar coordinate x=r\cos\theta

\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta

=3k\int_{0}^{\pi}\cos\theta d\theta

=3k\big[\sin\theta\big]_{0}^{\pi}

=3k\big[\sin\pi-\sin 0\big]

=0

Then, \bar{x}=\frac{M_y}{m}=0

Hence center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

3 0
3 years ago
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