Answer:
1. The third one
2. The first one
3.The fourth one
4. the first one
the thrid one
Step-by-step explanation:
Answer:
The correct answer is d.
Step-by-step explanation:
Let's solve each answer and see which one matches.
Note that in each case, you simply add or subtract from x inside the radical to move left or right respectively. Similarly, just add or subtract to the full expression to move up or down respectively. Doing so we get:
a) ![f(x) = \sqrt[3]{x + 7} + 3](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%20%2B%207%7D%20%2B%203)
b)![f(x) = \sqrt[3]{x - 3} - 7](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%20-%203%7D%20-%207)
c)![f(x) = \sqrt[3]{x - 3} + 7](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%20-%203%7D%20%2B%207)
d)![f(x) = \sqrt[3]{x - 7} + 3](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%20-%207%7D%20%2B%203)
So d is the correct answer.
By letting

we get derivatives


a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

Examine the lowest degree term
, which gives rise to the indicial equation,

with roots at r = 0 and r = 4/5.
b) The recurrence for the coefficients
is

so that with r = 4/5, the coefficients are governed by

c) Starting with
, we find


so that the first three terms of the solution are

If Heather puts in 1.36% of her salary into a retirement account and she makes $63,000 in a year, this can be solved by 1.36% x 63000. First convert 1.36% into a decimal. To do this, divide by 100.
1.36/100=0.0136
1.36% x 63000
=0.0136 x 63000
=856.8
Heather would put $856.8 in for the whole year. To find how much per month, divide by 12.
856.8/12=71.<span>4
</span>Rounded to the nearest dollar is $71.
Heather would put $71 into a retirement account each month.
<em>
</em><em>Note: This was solved by multiply the percent and then dividing by 12 months. You could also do this by dividing by 12 months first and then multiplying the percent.</em>
Answer:25th Percentile: 5
50th Percentile: 30
75th Percentile: 34
Interquartile Range: 29
Step-by-step explanation: