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DENIUS [597]
4 years ago
6

The average distance from Earth to the sun is about 9.3×10^7 miles. The average distance from Mars to the sun is about 1.4×10^8

miles. When both planets are at their average distance from the sun, how much farther is Mars from the sun than Earth?
Mathematics
1 answer:
diamong [38]4 years ago
5 0

Move the decimal to the right (you go to the right since the exponent is positive, if the exponent was negative then you would move left) seven times, giving you 9,300,000 miles (for Earth). You then do the same with the next equation, giving you 140,000,000 miles (for Mars).

Finally, to find the official number of miles, you simply work the equation 140,000,000 - 9,300,000 .. giving you 130,700,000 miles.


If your teacher wants you to put it back into exponential form, then you first make the number into a decimal that is greater than one and less than ten.


1.30700000


Then, what you want to do is count all of the zeros as a single exponent, giving you 5, but we're not done yet. To get the correct answer, you have to also count the numbers until you reach the decimal. Remember to take out the zeros, because they're not really necessary anymore!


1.307×10^{8}

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Todd earned some money doing chores. He spent one-fourth of his money see a movie. Then he spent $6.00 on popcorn and drinks. Wh
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Answer: Todd earned $40 mowing the neighbor's lawn.

Step-by-step explanation:

Let x represent the amount of money that Todd earned, mowing his neighbor's lawn.

He spent one-fourth of his money see a movie. It means that the amount spent in seeing a movie is x/4

Then he spent $6.00 on popcorn and drinks. It means that the total amount spent is

x/4 + 6

The amount left would be

x - (x/4 + 6) = x - x/4 - 6

When he went home, he had $24.00 left. It means that

x - x/4 - 6 = 24

x - x/4 = 24 + 6

x - x/4 = 30

Cross multiplying by 4, it becomes

4x - x = 120

3x = 120

x = 120/3

x = $40

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Step-by-step explanation:

4 0
2 years ago
Can someone help me with this? I need to find the points of discontinuity/limits for each of these. I think one point is 4, but
Debora [2.8K]
The answers are shown in the attached image

-------------------------------------------------------------------------

Explanation:

Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x

x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4

The x values 0 and 4 make the denominator zero

These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.

There are two vertical asymptotes

Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity

Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity

Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity

---------------------

Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity

Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side

So overall, as we approach x = 4, the y value is heading off to positive infinity

Again everything is summarized in the image attachment

Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation. 

6 0
3 years ago
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