All categories

3 years ago

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

Mathematics

1 answer:

mr_godi [17]3 years ago

7 0

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

You might be interested in

A card is drawn one at a time from a

ExtremeBDS [4]

Answer:

The experimental probability P(E) is 0.1818 = 18.18%

Step-by-step explanation:

The experimental probability is the number of desired outcomes divided by the number of total outcomes.

In this question:

11 trials(11 outcomes)

2 kings(2 desired)

P(E) = 2/11 = 0.1818

The experimental probability P(E) is 0.1818 = 18.18%

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cdisplaystyle%5Crm%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D%20%5

umka2103 [35]

Replace $x\mapsto \tan^{-1}(x)$ :

$\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \int_0^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx$

Split the integral at x = 1, and consider the latter one over [1, ∞) in which we replace $x\mapsto\frac1x$ :

$\displaystyle \int_1^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx = \int_0^1 \frac{\ln\left(\frac1x\right)}{\sqrt[3]{x} \left(1+\frac1{x^2}\right)} \frac{dx}{x^2} = - \int_0^1 \frac{\ln(x)}{\sqrt[3]{x} (1+x^2)} \, dx$

Then the original integral is equivalent to

$\displaystyle \int_0^1 \frac{\ln(x)}{1+x^2} \left(\sqrt[3]{x} - \frac1{\sqrt[3]{x}}\right) \, dx$

Recall that for |x| < 1,

$\displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x}$

so that we can expand the integrand, then interchange the sum and integral to get

$\displaystyle \sum_{n=0}^\infty (-1)^n \int_0^1 \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \ln(x) \, dx$

Integrate by parts, with

$u = \ln(x) \implies du = \dfrac{dx}x$

$du = \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \, dx \implies u = \dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac23}}{2n+\frac23}$

$\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \int_0^1 \left(\dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac13}}{2n-\frac13}\right) \, dx \\\\ = \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{\left(2n+\frac43\right)^2} - \frac1{\left(2n+\frac23\right)^2}\right) \\\\ = \frac94 \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$

Recall the Fourier series we used in an earlier question [27217075]; if $f(x)=\left(x-\frac12\right)^2$ where 0 ≤ x ≤ 1 is a periodic function, then

$\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi n x)}{n^2}$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(2\pi(3n+1)x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(2\pi(3n+2)x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(2\pi(3n)x)}{(3n)^2}\right)$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(6\pi n x + 2\pi x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(6\pi n x + 4\pi x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(6\pi n x)}{(3n)^2}\right)$

Evaluate f and its Fourier expansion at x = 1/2 :

$\displaystyle 0 = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+1)^2} + \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} + \sum_{n=1}^\infty \frac{(-1)^n}{(3n)^2}\right)$

$\implies \displaystyle -\frac{\pi^2}{12} - \frac19 \underbrace{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}}_{-\frac{\pi^2}{12}} = - \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$

$\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right) = \frac{2\pi^2}{27}$

So, we conclude that

$\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \frac94 \times \frac{2\pi^2}{27} = \boxed{\frac{\pi^2}6}$

3 0

2 years ago

Plzzzzz help QUICKKKKKKLY

guapka [62]

I believe the answer is,

${3x}^{5} + {7x}^{2} - {9x}^{4}$

3 0

3 years ago

Victoria is 4 years older than her neighbor. The sum of their ages is no more than 14 years.

marysya [2.9K]

Victoria is 9. Her neighbor is 5.

7 0

3 years ago

To earn money for this vacation, Grayson works at a local shop on weekends. His job is to cut bricks of fudge into 0.25 pound sq

postnew [5]

If each square brick cut is equal to .25 pounds and in total he cuts 36 equal sized squares, then you just need to multiply .25 by 36. One way to look at it is, what is 25% of 36. Whether or not you multiply by 25% or .25 the answer will be 9. Grayson has cut 9 pounds of fudge.

5 0

3 years ago