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schepotkina [342]
3 years ago
8

Find the percent of each number 0.9 of 1000

Mathematics
1 answer:
trasher [3.6K]3 years ago
8 0

We can use the is/of = p/100 method here. Our given values are 90% and 1000. Think: "What number is 90% of 1000?" Let's plug in our values.

x/1000 = 90/100

Solve for x.

x = (1000)(0.9)

x = 900

So, 900 is 90% of 1000.

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There are about 45000 farms in Nebraska using approximately 45 million acres of land. This farmland covers about 9/10 of the sta
yulyashka [42]

9/10 : 1/10 = 45,000,000 : x

x = 45000000/9 = 5,000,000

Answer: 5 million acres


6 0
3 years ago
You can pay $4.80 for 8 pounds of sugar or $11.75 for 25 pounds of
cricket20 [7]
25 pounds would be a better deal
7 0
3 years ago
Read 2 more answers
PLEASE HELP
natima [27]

Answer:

x= 5/8

Step-by-step explanation:

7 0
2 years ago
ABCD∼EFGH AD=45 in. , EH=75 in. , and AB=30 in. What is EF ? Enter your answer in the box. EF = in.
Helen [10]
The ~ symbol means that the 2 shapes are similar, meaning the same shape but not the same size.

To determine the measure of side length EF, you must find the relationship between the 2 side lengths.

You should set these up as ratios of the length to the width. So, for ABCD, it is 30/45. Make this equal to x/75. If you simplify the first ratio to 2:3, you can then multiply by both numerator and denominator by 25 to get x = 25 in/ 75 in.

Another way is to use cross products with you proportion. 75 x 30 = 45x. Solve for x to get 50 inches.
6 0
3 years ago
81 POINTS
Jobisdone [24]

Base Case: plug in n = 1 (the smallest positive integer)

If n = 1, then 3n-2 = 3*1-2 = 1. Square this and we see that (3n-2)^2 = 1^2 = 1

On the right hand side, plugging in n = 1 leads to...

n*(6n^2-3n-1)/2 = 1*(6*1^2-3*1-1)/2 = 1

Both sides are 1. So that confirms the base case.

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

Inductive Step: Assume that

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

is a true statement for some positive integer k. If we can show the statement leads to the (k+1)th case being true as well, then we will have sufficiently proven the overall statement to be true by induction.

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 + (3(k+1)-2)^2 = (k+1)*(6(k+1)^2-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3(k+1)-2)^2 = (k+1)*(6(k^2+2k+1)-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3k+3-2)^2 = (k+1)*(6k^2+12k+6-3k-3-1)/2

k*(6k^2-3k-1)/2 + (3k+1)^2 = (k+1)*(6k^2+9k+2)/2

k*(6k^2-3k-1)/2 + 9k^2+6k+1 = (k+1)*(6k^2+9k+2)/2

(6k^3-3k^2-k)/2 + 2(9k^2+6k+1)/2 = (k*(6k^2+9k+2)+1(6k^2+9k+2))/2

(6k^3-3k^2-k + 2(9k^2+6k+1))/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3-3k^2-k + 18k^2+12k+2)/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3+15k^2+11k+2)/2 = (6k^3+15k^2+11k+2)/2

Both sides simplify to the same expression, so that proves the (k+1)th case immediately follows from the kth case

That wraps up the inductive step. The full induction proof is done at this point.

7 0
3 years ago
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