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fgiga [73]
3 years ago
10

Marry Allen baked 15 cupcakes and her children ate 2/3 of them. How many cupcakes did they eat?

Mathematics
1 answer:
ehidna [41]3 years ago
3 0

Answer:

10 were eaten

Step-by-step explanation:

If we have 15 cupcakes and 2/3 of them were eaten, we multiply 15 times 2/3

15*2/3

12*2 = 30

30/3 = 10

10 of the cupcakes were eaten

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The base of a solid right pyramid is a square with an
zhuklara [117]

Answer:

Volume = ⅓n²(n-1) or ⅓(n³ - n²)

Step-by-step explanation:

Given

Solid Shape: Right pyramid

Edge= n units

Height= n - 1 units

Required

Volume of the pyramid

The volume of a right pyramid is

Volume = ⅓Ah

Where A represents the area of the base

h represent the height of the pyramid

Since it has a square base;

The area is calculated as follows

Area, A = edge * edge

A = n * n

A = n²

Recall that

Volume = ⅓Ah

Substitute n² for A and n - 1 for h

The expression becomes

Volume = ⅓ * n² * (n - 1)

Volume = ⅓n²(n-1)

The expression can be solved further by opening the bracket

Volume = ⅓(n³ - n²)

6 0
3 years ago
What is the slope and y-intercept of this equation? y=1/2 x - 6
LenKa [72]

Answer:

hi im bob

Step-by-step explanation:

8 0
3 years ago
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If you roll a six-sided die 6 times what is the best prediction possible for the number okay you were Roll a 2
stepladder [879]
6 i think im not sure

3 0
4 years ago
Read 2 more answers
What is the value of x in the diagram below?
VARVARA [1.3K]

Answer:

<h2>7.2</h2>

option B is the right option.

Step-by-step explanation:

<h3>Using leg rule</h3>

\frac{bc}{ab}  =  \frac{ab}{bd}

Plug the values:

\frac{20}{12}  =  \frac{12}{x}

Apply cross product property

20 \times x = 12 \times 12

Calculate the product

20x = 144

divide both sides of the equation by 20

\frac{20x}{20}  =  \frac{144}{20}

Calculate:

x = 7.2

hope this helps..

Good luck...

4 0
3 years ago
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Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
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