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nevsk [136]
3 years ago
11

There are 120 Calories in a 3/4-cup serving of cereal. How many Calories are there in 6 cups of cereal? Enter your answer in the

box.
Mathematics
2 answers:
DiKsa [7]3 years ago
5 0
If 3/4 of a cup is 120 calories, then 1/4 would be 40 calories. To see how many calories there are in a full cup, we would multiply 40 calories by 4 to get a full cup. 40 * 4 = 160. So, there are 160 calories in 1 cup of cereal. To see how many calories there are in 6 cups, we simply multiply 160 by 6. 160 * 6 = 960.

So, there are 960 calories in 6 cups of cereal.


Pie3 years ago
5 0

Answer:

there are 960 calories in 6 cups of cereal.

Step-by-step explanation:

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There were 3232 volunteers to donate blood. Unfortunately, nn of the volunteers did not meet the health requirements, so they co
Degger [83]

Answer:

\frac{470}{32-n}  

Step-by-step explanation:  

We have been given that there were 32 volunteers to donate blood. Unfortunately, n of the volunteers did not meet the health requirements, so they couldn't donate.            

So the number of volunteers that donated blood will be 32-n.

We are also told that the rest of the volunteers donated 470 milliliters each.

To find the units of blood donated by each of the volunteers we will divide total units of donated blood by number of volunteers, who donated the blood (32-n).

\text{Millimeters of blood donated by each volunteer}=\frac{470}{32-n}  

Therefore, each of the volunteers donated \frac{470}{32-n} millimeters of blood.  

5 0
2 years ago
Read 2 more answers
What are the angle measures of triangle ABC?
Schach [20]

The angle measures of triangle ABC is ∠A=90°, ∠B= 60°, ∠C = 30°. Thus, the correct option is B.

<h3>What is a triangle?</h3>

A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The sum of all the angles of a triangle is always equal to 180°.

As it is known that the angles of the triangle are 30°, 60°, and 90°. Therefore, the largest angle is 90° which will be the opposite of the largest side of the triangle, therefore, the measure of ∠A=90°.

Also, the shortest angle of the triangle is 30°, which will lie at the opposite of the smallest side of the triangle, therefore, the measurement of the ∠C=30°.

Now, the angle left is ∠B therefore, the measure of ∠B=60°.

Hence, the angle measures of triangle ABC is ∠A=90°, ∠B= 60°, ∠C = 30°. Thus, the correct option is B.

Learn more about Triangle:

brainly.com/question/2773823

#SPJ1

4 0
2 years ago
Fortninte.com had 595,760 visitors in June 2018, and 615,355 in July 2018. How many more visitors came in July than June?
Inessa [10]

Solve. Subtract July 2018 with June 2018

615355 - 595760 = 19595

19595 more people visited Fortnite's website in July than June.

hope this helps

<em>~Rise Above the Ordinary</em>


6 0
2 years ago
Read 2 more answers
A prism and two nets are shown below:
nikdorinn [45]
Part a : Net A because when you fold it up it will look like the wanted prism

Part b : AB = 3 in
BC = 5 in
CD = 7.2 in

Part c : area of one triangle = 4*3 = 12 / 2 = 6
6 * 2 = 12 (because there are 2 triangles)
7.2 * 3 = 21.6
7.2 * 4 = 28.8
7.2 * 5 = 36
Add them all up : 12+21.6+28.8+36 = 98.4
So surface area is 98.4 in
4 0
2 years ago
CALC- limits<br> please show your method
gladu [14]
A. Factor the numerator as a difference of squares:

\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6

c. As x\to\infty, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4

e. Let's first rewrite the root terms with rational exponents:

\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}

Next we rationalize the numerator and denominator. We do so by recalling

(a-b)(a+b)=a^2-b^2
(a-b)(a^2+ab+b^2)=a^3-b^3

In particular,

(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3
(x^{1/2}-x)(x^{1/2}+x)=x-x^2

so we have

\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}

For x\neq0 and x\neq1, we can simplify the first term:

\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x

So our limit becomes

\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43
3 0
3 years ago
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