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3 years ago

Given a cube with a volume of 27 cm3, what is the volume of a square pyramid that can fit perfectly inside the cube?

Mathematics

2 answers:

Artist 52 [7]3 years ago

6 0

It's nine. You rom FLVS too?

Arturiano [62]3 years ago

5 0

Answer:

9cm3

Correct on quiz

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Which equation has the steepest graph?

prisoha [69]

C. Y = -10x -4 as this is the largest slope with a slope of -10/1

5 0

3 years ago

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Julian walked 6/10 of a mile to his friends house and another 35/100 mile to the store. He walked 1/4 of a mile back home. Julia

AveGali [126]

The statement, "Julian's sister said he walked 1/5 mile" cannot be agreed because Julian totally walked $1\frac{1}{5} \text{ or } \frac{6}{5}$ miles.

<u>Solution:</u>

Given that,

Julian walked 6/10 of a mile to his friends house
Another 35/100 mile to the store
He walked 1/4 of a mile back home

To find total distance walked by Julian we have to add the above stated values. That is, $\frac{6}{10} +\frac{35}{100} +\frac{1}{4}$

Factors of 10 = $5\times2$

Factors of 100 = $5\times2\times5\times2$

Factors of 4 = $2\times2$

Therefore, the least common factor of 10, 100 and 4 is 100. With like denominators we can operate on just the numerators,

$\frac{6\times10}{10\times10} +\frac{35\times1}{100\times1} +\frac{1\times25}{4\times1}\rightarrow\frac{60+35+25}{100}\rightarrow\frac{120}{100}$

$\Rightarrow\frac{120}{100}\rightarrow\frac{6}{5}$

Which can also be written as $1\frac{1}{5}$ .

So, from the above calculation it can be said that Julian walked $1\frac{1}{5} \text{ miles }$ .

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3 years ago

Write an expression for the sequence of operations described below.

melisa1 [442]

T to the 4th power then multiply by 3.

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3 years ago

Write a unit rate for the situation.

tamaranim1 [39]

Answer:

50 dollors

Step-by-step explanation:

8 0

2 years ago

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Give an example of a function f : N → N that is surjective but not injective. You must explain why your example is surjective an

Helen [10]

Answer:

Consider f: N → N defined by f(0)=0 and f(n)=n-1 for all n>0.

Step-by-step explanation:

First we will prove that f is surjective. Let y∈N be any natural number. Define x as the number x=y+1. Then x∈N, and f(x)=x-1=(y+1)-1=y. We conclude that f is surjective.

However, f is not injective. Take x1=0 and x2=1. Then x1≠x2 but f(x1)=0 and f(x2)=x2-1=1-1=0. We have shown that there are two natural numbers x1,x2 such that x1≠x2 but f(x1)=f(x2), that is, f is not injective.

Note:

If 0∉N in your definition of natural numbers, the same reasoning works with the function f: N → N defined by f(1)=1 and f(n)=n-1 for all n>1. The only difference is that you consider x1=1, x2=2 for the injectivity.

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2 years ago