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Tresset [83]
4 years ago
7

Jermaine wants to know how fast he is traveling in a school bus, but he cannot see the speedometer. He records that the bus trav

els 1 mile in 75 seconds. How fast is Jermaine traveling in miles per hour?
Mathematics
1 answer:
Advocard [28]4 years ago
8 0
<h3>Answer:</h3>

48 mph

<h3>Explanation:</h3>

The rate in miles per second can be converted to the rate in miles per hour by multiplying by the number of seconds in an hour:

\dfrac{1\,mi}{75\,s}\cdot\dfrac{3600\,s}{1\,h}=\dfrac{3600\,mi}{75\,h}=48\,mi/h

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Aleks04 [339]
-23 is a negative number so that means 7 is a greater number. any number with a minus sign like this - means it’s a negative number, so any number that you have to compare it to and if that number has nothing before or after it, then that would be the greater number
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Gina has 21 more barrettes than Holly. The equation g = 21 + h, where g represents the number of barrettes Gina has, and h repre
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Answer:

21

Step-by-step explanation:

42-21=21

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Mario had 9 green 810 Brown 6 Orange and 9 blue m&amp;m's what is the fractions of M&amp;M's that are orange? Help please
Ipatiy [6.2K]
Hey You!

Let's add up the total about of M&M's.

9 + 810 + 6 + 9 = 834


That is the denominator for the fraction. The numerator is the amount of orange M&M's, which have a quantity of 6.

= 6/834

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3 years ago
If Sarah has 30 bills in her wallet worth $240, all fives and tens, how many of each bill does she have? Show work if possible
bonufazy [111]

Answer:

Step-by-step explanation:

3 0
3 years ago
I need help please. Thanks!
Karolina [17]

Answer:

A

Step-by-step explanation:

We are given the function f and its derivative, given by:

f^\prime(x)=x^2-a^2=(x-a)(x+a)

Remember that f(x) is decreasing when f'(x) < 0.

And f(x) is increasing when f'(x) > 0.

Firstly, determining our zeros for f'(x), we see that:

0=(x-a)(x+a)\Rightarrow x=a, -a

Since a is a (non-zero) positive constant, -a is negative.

We can create the following number line:

<-----(-a)-----0-----(a)----->

Next, we will test values to the left of -a by using (-a - 1). So:

f^\prime(-a-1)=(-a-1-a)(-a-1+a)=(-2a-1)(-1)=2a+1

Since a is a positive constant, (2a + 1) will be positive as well.

So, since f'(x) > 0 for x < -a, f(x) increases for all x < -a.

To test values between -a and a, we can use 0. Hence:

f^\prime(0)=(0-a)(0+a)=-a^2

This will always be negative.

So, since f'(x) < 0 for -a < x < a, f(x) decreases for all -a < x < a.

Lasting, we can test all values greater than a by using (a + 1). So:

f^\prime(a+1)=(a+1-a)(a+1+a)=(1)(2a+1)=2a+1

Again, since a > 0, (2a + 1) will always be positive.

So, since f'(x) > 0 for x > a, f(x) increases for all x > a.

The answer choices ask for the domain for which f(x) is decreasing.

f(x) is decreasing for -a < x < a since f'(x) < 0 for -a < x < a.

So, the correct answer is A.

3 0
3 years ago
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