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3 years ago

There are 12 grams of sugar in 1⁄3 of a piece of candy. How much sugar is in 3⁄4 of a piece of candy?

Mathematics

1 answer:

lisov135 [29]3 years ago

4 0

Answer:

27 grams of sugar

Step-by-step explanation:

If 1/3 candy contains 12 grams of sugar

Then 3/4 candy contains C grams of sugar

C= 3/4 * 12grams➗1/3=3*12*3grams➗4=108grams➗4=27 grams of sugar

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Do you want to $140 to your friend he pays you back $270 what interest rate did you charge your friend

ExtremeBDS [4]

Answer:

the interest rate is 92.86%

Step-by-step explanation:

270-140= 130

the %interest= (130/140)×100%

92.86%

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3 years ago

50 POINTS what is the average rate of change of the function f(x)=3x^2 on the interval from 0 to 2?

Evgesh-ka [11]

Answer:

f=3x-2

Step-by-step explanation:

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3 years ago

A group of engineers is building a parabolic satellite dish whose shape will be formed by rotating the curve y=ax2 about the y-a

Blizzard [7]

Answer:

Step-by-step explanation:

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3 years ago

Suppose that a large pumpkin empty a swimming pool in 40 hours and that a small pumpkin empty a pool and 70 hours working togeth

Norma-Jean [14]

Answer:

Total 25.45 hours will be taken to empty the pool.

Step-by-step explanation:

Time taken by large pumpkin to empty the pool=40 hours

Time taken by the smaller pumpkins to empty the pool=70 hours

Amount of pool emptied by large pumpkin in one hour= $\frac{1}{40}$

Amount of pool emptied by small pumpkin in one hour= $\frac{1}{70}$

Amount of pool emptied in one hour if both pumpkins work together=

$\frac{1}{40}$ + $\frac{1}{70}$

= $\frac{11}{280}$

Therefore, the no. of days will be the reciprocal of the amount of work done in one hour ie . $\frac{280}{11}$

Total 25.45 hours will be taken to empty the pool.

5 0

3 years ago

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much

Y_Kistochka [10]

Answer:

x = ∛ 2*V/5

y = ∛ 2*V/5

h = V/ ∛ 4*V²/25

Step-by-step explanation:

Dimensions of the aquarium base is x*y

We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.

let call h the height of the aquarium then volume of the aquarium is:

V = x*y*h where h = V / x*y

As the base is a rectangular one there are 2 sides x*h . and 2 sides y*h

According to this:

Ct (cost of aquarium ) = cost of the base + cost of the sides

cₐ ( cost of the base) = 5*c₁*x*y

c₆ (cost of the sides ) = c₁*2*x*h + c₁*2*y*h

C(t) = 5*c₁*x*y +2* c₁*x* V/x*y + 2* c₁*y* V/x*y or

C(t) = 5*c₁*x*y + 2*c₁*V/y *2*c₁* V/x

Taking partial derivatives en x and y we have:

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(y) = 5*c₁*x - 2*c₁*V/y²

C´(x) = C´(y) ⇒ 5*c₁*y - 2*c₁*V/x² = 5*c₁*x - 2*c₁*V/y²

or 5*y - 2*V/x² = 5*x - 2*V/y²

(5*y*x² - 2*V)/x² = ( 5*y²x - 2*V) /y²

(5*y*x² - 2*V)*y² = ( 5*y²x - 2*V)*x²

5*y³*x² - 2*V*y² = 5*y²x³ - 2*V*x²

5*y³*x² - 5*y²x³ = 2*V * ( y² - x²)

by symmetry x = y

Then using x = y and plugging that value on the derivatives

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(x) = 5*c₁*x - 2*c₁*V/x²

C´(x) = 0 ⇒ 5*c₁*x - 2*c₁*V/x² = 0

5*x - 2*V/x² = 0 ⇒ 5*x³ - 2*V = 0 ⇒ 5*x³ = 2*V ⇒ x³ = 2*V/5

x = ∛ 2*V/5 and y = ∛ 2*V/5 and h = V/ x*y h = V/ ∛ 4*V²/25

7 0

3 years ago