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3 years ago

15

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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Helpp me with this word problem number one

Ymorist [56]

In order to find out how long it is you need to convert the fractions into mixed fractions then multiply.
4 1/3 x 6 1/2 =
13/3 x 13/2 = 169/6
169 divided by 6 = 28 1/6

6 0

3 years ago

Find dy/dx x^3+y^3=18xy

tatyana61 [14]

Differentiate both sides of the equation.<span><span><span>d<span>dx</span></span><span>(<span>x3</span>+<span>y3</span>)</span>=<span>d<span>dx</span></span><span>(18xy)</span></span><span><span>d<span>dx</span></span><span>(<span>x3</span>+<span>y3</span>)</span>=<span>d<span>dx</span></span><span>(18xy)</span></span></span>Differentiate the left side of the equation.Tap for fewer steps...By the Sum Rule, the derivative of <span><span><span>x3</span>+<span>y3</span></span><span><span>x3</span>+<span>y3</span></span></span> with respect to <span>xx</span> is <span><span><span>d<span>dx</span></span><span>[<span>x3</span>]</span>+<span>d<span>dx</span></span><span>[<span>y3</span>]</span></span><span><span>d<span>dx</span></span><span>[<span>x3</span>]</span>+<span>d<span>dx</span></span><span>[<span>y3</span>]</span></span></span>.<span><span><span>d<span>dx</span></span><span>[<span>x3</span>]</span>+<span>d<span>dx</span></span><span>[<span>y3</span>]</span></span><span><span>d<span>dx</span></span><span>[<span>x3</span>]</span>+<span>d<span>dx</span></span><span>[<span>y3</span>]</span></span></span>Differentiate using the Power Rule which states that <span><span><span>d<span>dx</span></span><span>[<span>xn</span>]</span></span><span><span>d<span>dx</span></span><span>[<span>xn</span>]</span></span></span> is <span><span>n<span>x<span>n−1</span></span></span><span>n<span>x<span>n-1</span></span></span></span> where <span><span>n=3</span><span>n=3</span></span>.<span><span>3<span>x2</span>+<span>d<span>dx</span></span><span>[<span>y3</span>]</span></span><span>3<span>x2</span>+<span>d<span>dx</span></span><span>[<span>y3</span>]</span></span></span>Evaluate <span><span><span>d<span>dx</span></span><span>[<span>y3</span>]</span></span><span><span>d<span>dx</span></span><span>[<span>y3</span>]</span></span></span>.Tap for more steps...<span><span>3<span>x2</span>+3<span>y2</span><span>d<span>dx</span></span><span>[y]</span></span><span>3<span>x2</span>+3<span>y2</span><span>d<span>dx</span></span><span>[y]</span></span></span>Differentiate the right side of the equation.Tap for fewer steps...Since <span>1818</span> is constant with respect to <span>xx</span>, the derivative of <span><span>18xy</span><span>18xy</span></span> with respect to <span>xx</span> is <span><span>18<span>d<span>dx</span></span><span>[xy]</span></span><span>18<span>d<span>dx</span></span><span>[xy]</span></span></span>.<span><span>18<span>d<span>dx</span></span><span>[xy]</span></span><span>18<span>d<span>dx</span></span><span>[xy]</span></span></span>Differentiate using the Product Rule which states that <span><span><span>d<span>dx</span></span><span>[f<span>(x)</span>g<span>(x)</span>]</span></span><span><span>d<span>dx</span></span><span>[f<span>(x)</span>g<span>(x)</span>]</span></span></span> is <span><span>f<span>(x)</span><span>d<span>dx</span></span><span>[g<span>(x)</span>]</span>+g<span>(x)</span><span>d<span>dx</span></span><span>[f<span>(x)</span>]</span></span><span>f<span>(x)</span><span>d<span>dx</span></span><span>[g<span>(x)</span>]</span>+g<span>(x)</span><span>d<span>dx</span></span><span>[f<span>(x)</span>]</span></span></span> where <span><span>f<span>(x)</span>=x</span><span>f<span>(x)</span>=x</span></span> and 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<span><span><span>y2</span>−6x</span><span><span>y2</span>-6x</span></span>.<span><span><span><span>3y'<span>(<span>y2</span>−6x)</span></span><span><span>y2</span>−6x</span></span>=−<span><span>3<span>x2</span></span><span><span>y2</span>−6x</span></span>+<span><span>18y</span><span><span>y2</span>−6x</span></span></span><span><span><span>3y′<span>(<span>y2</span>-6x)</span></span><span><span>y2</span>-6x</span></span>=-<span><span>3<span>x2</span></span><span><span>y2</span>-6x</span></span>+<span><span>18y</span><span><span>y2</span>-6x</span></span></span></span>Reduce the expression by cancelling the common factors.Tap for more steps...<span><span>3y'=−<span><span>3<span>x2</span></span><span><span>y2</span>−6x</span></span>+<span><span>18y</span><span><span>y2</span>−6x</span></span></span><span>3y′=-<span><span>3<span>x2</span></span><span><span>y2</span>-6x</span></span>+<span><span>18y</span><span><span>y2</span>-6x</span></span></span></span>Simplify the right side of the equation.Tap for more steps...<span><span>3y'=−<span><span>3<span>(<span>x2</span>−6y)</span></span><span><span>y2</span>−6x</span></span></span><span>3y′=-<span><span>3<span>(<span>x2</span>-6y)</span></span><span><span>y2</span>-6x</span></span></span></span>Divide each term by <span>33</span> and simplify.Tap for fewer steps...Divide each term in <span><span>3y'=−<span><span>3<span>(<span>x2</span>−6y)</span></span><span><span>y2</span>−6x</span></span></span><span>3y′=-<span><span>3<span>(<span>x2</span>-6y)</span></span><span><span>y2</span>-6x</span></span></span></span> by <span>33</span>.<span><span><span><span>3y'</span>3</span>=−<span><span><span>3<span>(<span>x2</span>−6y)</span></span><span><span>y2</span>−6x</span></span>3</span></span><span><span><span>3y′</span>3</span>=-<span><span><span>3<span>(<span>x2</span>-6y)</span></span><span><span>y2</span>-6x</span></span>3</span></span></span>Reduce the expression by cancelling the common factors.Tap for more steps...<span><span>y'=−<span><span><span>3<span>(<span>x2</span>−6y)</span></span><span><span>y2</span>−6x</span></span>3</span></span><span>y′=-<span><span><span>3<span>(<span>x2</span>-6y)</span></span><span><span>y2</span>-6x</span></span>3</span></span></span>Simplify the right side of the equation.Tap for more steps...<span><span>y'=−<span><span><span>x2</span>−6y</span><span><span>y2</span>−6x</span></span></span><span>y′=-<span><span><span>x2</span>-6y</span><span><span>y2</span>-6x</span></span></span></span>Replace <span><span>y'</span><span>y′</span></span> with <span><span><span>dy</span><span>dx</span></span><span><span>dy</span><span>dx</span></span></span>.<span><span><span>dy</span><span>dx</span></span>=−<span><span><span><span>x2</span>−6y</span><span><span>y2</span>−6x</span></span></span></span>

6 0

3 years ago

I need help because I'm solving these questions by the manual, and yet it still says that I am wrong after multiple attempts.

Alexxandr [17]

Answer:

1. QR = HI

2. angle W and angle T

3. WB = UV

Step-by-step explanation:

It could me fill in the part of the proof, since I notice on all of them they only mark one or two of the congruencies on the shapes. I'm not really sure of 3's answer.

Hope this helped somewhat and good luck

6 0

3 years ago

Find f'(x) for f(x) = 5x2 − 3x + 12. 10x + 12(this answer is incorrect) −10x − 3 10x − 3 None of these

nexus9112 [7]

What are ya sayin

Step-by-step explanation:

5 0

3 years ago

How to use the Pythagorean theorem to provee if a triangle with the side lengths 7,14,28 is a right triangle. Plz help

Lapatulllka [165]

Answer: find the answer in the explanation.

Step-by-step explanation:

To use the Pythagorean theorem to prove if a triangle with the side lengths 7,14,28 is a right triangle, you will square 7 and add it to the square of 14. The square root of the sum will be equal to 28 if it is a right angle triangle. Otherwise, it is not.

(7^2 + 14^2) = 49 + 196

245

Find the square root of 245

Square root = 15.65

Since it is not 24, the triangle is not a right angle triangle.

8 0

3 years ago