Simplify (7xy)^2 ????

A bakery uses ounces of icing for every of a cake. What is the unit rate in ounces of icing per cake?

mr_godi [17]

Answer:

The unit rate in ounces of icing per cake is 40 4/5 or 204/5

Complete statement and question:

A bakery uses 10 1/5 ounces of icing for every 1/4 of a cake. What is the unit rate in ounces of icing per cake

Source: A previous question found at brainly

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Size of icing for every 1/4 of a cake = 10 1/5 ounces or 51/5 ounces

2. What is the unit rate in ounces of icing per cake?

For answering the question, we will use the Rule of Three Simple, this way:

Unit rate in ounces of icing per cake * 1/4 of a cake = Total cake * Unit rate in ounces of icing per 1/4 of a cake

Replacing with the real values, we have:

Unit rate in ounces of icing per cake * 1/4 = 1 * 51/5

Unit rate in ounces of icing per cake = 51/5 / 1/4

Unit rate in ounces of icing per cake = 51/5 * 4/1

Unit rate in ounces of icing per cake = 204/5 ounces or 40 4/5 ounces

The unit rate in ounces of icing per cake is 40 4/5 or 204/5

7 0

3 years ago

What are the best strategies to compare fractions. Name as many strategies as you want.

inn [45]

Common denominator
Common multiple
Simplest Form
GCF: Greatest Common Factor
LCM: Least Common Factor
Then Comparing Plain

4 0

3 years ago

Read 2 more answers

Abigail redecorates her house. A scale drawing of her house can be seen above. If each 6cm on the scale drawing equals 12ft, wha

mihalych1998 [28]

If 6cm=12ft, then if you reduce that you get 1cm=2ft so multiply each measurement by 2 getting 9cm=18ft, 7cm=14ft, 10cm=20ft, 8cm=16ft

7 0

3 years ago

A football team loses 3 yards in 3 consecutive plays. What’s the total yardage gained?

Anni [7]

-3×3=-9 so gain is 0

4 0

3 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the bin is constructed from 48 square feet of sheet metal, so:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

3 years ago