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3 years ago

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Jessie is following a chocolate milk recipe that requires of 2 cups milk and 5 tablespoons of cocoa powder. He wants to make a l

arge batch. How many cups of milk does he need if the batch requires 40 tablespoons of cocoa powder?

1 answer:

Reika [66]3 years ago

3 0

Step-by-step explanation:

5 tablespoons = 2 cups milk

1 tablespoon = 2/5 cups milk

40 tablespoons = 2/5 × 40 = 16 cups milk

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Solve 5w + 3 = 3w + 10

OverLord2011 [107]

X=7/12
Alright so let's solve this! :)
••• Subtract 3w from both sides.
<span><span><span><span>5w</span>+3</span>−<span>3w</span></span>=<span><span><span>3w</span>+10</span>−<span>3w</span></span></span><span><span><span>2w</span>+3</span>=10
</span>••• Subtract 3 from both sides.
<span><span><span><span>2w</span>+3</span>−3</span>=<span>10−3</span></span><span><span>2w</span>=7
</span>••• Divide both sides by 2.
<span>2w/2</span>=<span>7/<span>2
</span></span>=<span>7/2
and then 7/2=3.5</span>

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3 years ago

Read 2 more answers

Find the moment of inertia about the y-axis of the thin semicirular region of constant density

arlik [135]

The moment of inertia about the y-axis of the thin semicircular region of constant density is given below.

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

<h3>What is rotational inertia?</h3>

Any item that can be turned has rotational inertia as a quality. It's a scalar value that indicates how complex it is to adjust an object's rotational velocity around a certain axis.

Then the moment of inertia about the y-axis of the thin semicircular region of constant density will be

$\rm I_x = \int y^2 dA\\\\I_y = \int x^2 dA$

x = r cos θ

y = r sin θ

dA = r dr dθ

Then the moment of inertia about the x-axis will be

$\rm I_x = \int _0^r \int _0^{\pi} (r\sin \theta )^2 \ r \ dr \ d\theta\\\\\rm I_x = \int _0^r \int _0^{\pi} r^3 \sin ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_x = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about the y-axis will be

$\rm I_y = \int _0^r \int _0^{\pi} (r\cos\theta )^2 \ r \ dr \ d\theta\\\\\rm I_y = \int _0^r \int _0^{\pi} r^3 \cos ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about O will be

$\rm I_o = I_x + I_y\\\\I_o = \dfrac{1}{8} \times \pi r^4 + \dfrac{1}{8} \times \pi r^4\\\\I_o = \dfrac{1}{4} \times \pi r^4$

More about the rotational inertia link is given below.

brainly.com/question/22513079

#SPJ4

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2 years ago

Help me please thank you

MariettaO [177]

(x+4)+135.8=180.0
-135.8.-135.8

x+4= 44.2
-4 -4
x= 40.2

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3 years ago

A restaurant keeps track of sales each week. According to the scatter plot, which amount is an estimate of sales that will be ma

Ksenya-84 [330]

Answer:

it should be B

Step-by-step explanation:

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2 years ago

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In three basketball games over a weekend, 125,429 people came to watch. The next weekend , 86,353 people came to watch the games

Tems11 [23]

125,429
+ 86,353
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211,845

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3 years ago