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3 years ago

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Given the reaction at equilibrium:

1 answer:

AVprozaik [17]3 years ago

8 0

I think answer is a. Please give me brainlest I hope this helps let me know if it’s correct or not

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Which ratio is equivalent to 2/5 with greater terms

Dominik [7]

1.2 I’m so sorry if it’s wrong.

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3 years ago

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What is the value of X?

11111nata11111 [884]

Answer:

56

Step-by-step explanation:

You add 91 and 33 and you get 124. Subtract 180 and 124 to get 56.

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3 years ago

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What is the solution to the system of equations? you must show all work to recieve full credit!

earnstyle [38]

Use the substitution method:
2x - 6y = -12
x - 2y = -8
Then x = 2y - 8
Substitute in the first equation:
2(2y - 8) - 6y = -12
4y - 16 - 6y = -12
-2y = 4
y = -2
Now substitute y in one of the two equations given you prefer:
For example x-2*(-2) = -8
x + 4 = -8
x = -12
The solutions are x = -12 and y = -2

5 0

3 years ago

Leah is having a brunch. She wants to serve her guests 2 gallons of juice that is 75% orange juice and 25% pineapple juice. She

Katen [24]

She wants to serve --------- > 2 gallons of juice that is 75% orange juice and 25% pineapple juice

then

2*0.75------------------ > 1.5 gallons of orange juice

2*0.25------------------ > 0.5 gallons of pineapple juice

2 gallons------------ >1.5 gallons of orange juice+ 0.5 gallons of pineapple juice

if 1 gallons pineapple juice mixture--------------------- > 0.40 gallons pineapple juice

X------------------------------------------------------------------- > 0.50 gallons pineapple juice

X=50/40=1.25 gallons juice mixture

1.25 mixture gallons---- > 0.50 gallons pineapple juice+0.75 gallons orange juice

Therefore

(2-1.25)=0.75 gallons of orange juice

2 gallons------------ >0.75 gallons of orange juice+ <span>1.25 mixture gallons </span>

0.75*(p)+1.25*(m)=2--------------- > (0.75/1.25)*(p)+(1.25/1.25)*m=2/1.25

0.60p+m=1.6

The answer is 0.60p+m=1.6

5 0

3 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