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3 years ago

Jackson bought 3 pounds of candies for $9.75. c What was the price of these candies in cents per ounce?

2 answers:

7 0

Jackson bought 3*16 = 48 ounces for 975 cents. The unit price in cents per ounce is ...
975 cents/(48 oz) = 20.3125 cents/oz

Anon25 [30]3 years ago

4 0

Answer: The price of candies in cents per ounce is 20.31 cents per ounce.

Step-by-step explanation: Given that Jackson bought 3 pounds of candies for $9.75.

We are to find the price of the candies in cents per ounce.

We know that

1 dollar = 100 cents and 1 pound = 16 ounces.

Given that

Price of 3 pounds of candies = $ 9.75 = 9.75 × 100 cents = 975 cents.

So, price of 1 pound of candies will be

$\dfrac{975}{3}=325~\textup{cents}.$

This implies that, price of 16 ounces of candies = 325 cents.

Therefore, price of 1 ounce of candies will be

$\dfrac{325}{16}=20.31~\textup{cents}.$

Thus, the price of candies in cents per ounce is 20.31 cents per ounce.

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Which choices are equivalent to the fraction below check all the apply ""24" over 30

user100 [1]

Answer:

c) The reduced form of the given fraction $\frac{24}{30} = \frac{4}{5}$

Step-by-step explanation:

Here, the given expression is "24 over 30".

The given expression is is equivalent to $\frac{24}{30}$

Now by Prime Factorization:

24 = 2 x 2 x 2 x 3

30 = 2 x 3 x 5

⇒ The common factors in 24 and 30 is 2 x 3 = 6

So, $\frac{24}{30} = \frac{2 \times 2 \times 2 \times 2 \times 3}{2 \times 3 \times 5} = \frac{6 \times 4}{6 \times 5} = \frac{4}{5}$

Hence the reduced form of the given fraction $\frac{24}{30} = \frac{4}{5}$

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3 years ago

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

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Vlada [557]

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2 years ago

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Troyanec [42]

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Step-by-step explanation:

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