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2 years ago

Is 45 greater than 45 1/2

Mathematics

2 answers:

ioda2 years ago

6 0

Answer:

NO not at all, 45 1/2 is a half greater than 45

Step-by-step explanation:

Convert the fraction to a decimal to make it easier

45 1/2 is the same as 45.5 which as you can see is greater than 45

45.5 > 45

Hope this helps

- Benny

tresset_1 [31]2 years ago

5 0

Answer:

Step-by-step explanation:

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aliya0001 [1]

Hey!

This could be rewritten as:

2x + 1 + 5x

We can add like terms together:

7x + 1

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The number of people that belong to a certain social website is 412 and growing at a rate of 5% every month. How many members wi

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It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportat

katrin2010 [14]

Answer: $H_0:\mu=120$

$H_A:\mu\neq120$

Step-by-step explanation:

Let $\mu$ denotes the average braking distance for a population of small car.

Given : It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet.

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Objective of test : Whether the statement made in the advertisement is false

i.e. $\mu\neq120$

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So the null and the alternative hypotheses for the test will be:

$H_0:\mu=120$

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3 years ago

Find a fraction with a demominator of 30 that is equivalent to 1/3

Harman [31]

Answer:

10/30

Step-by-step explanation:

First, multiply the denominator by 10 to make 30.

Then, since we multiplied the denominator, we multiply the numerator. 1 x 10= 10

Therefore the answer is 10/30.

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2 years ago

F(x)=X over x^3-2x^2+5x why will this have no zeros

Mumz [18]

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

$\dfrac{x}{x^3-2x^2+5x}=0\iff x=0$

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

$\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}$

So, we can't even extend with continuity this function in such a way that $f(0)=0$

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3 years ago