Answer:
$2000
Step-by-step explanation:
let x be the money she invested
lets assume this was for 1 year
0.09(x/2) + 0.07(x/2) = 160
multiply each side by 2 to cancel the denominators:
0.09x + 0.07x = 320
0.16x = 320
x = 2000
The sum of all the even integers between 99 and 301 is 20200
To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.
In this case, the first even integer will be 100 and the last even integer will be 300.
nth term of the AP = first term + (n-1) x common difference
⇒ 300 = 100 + (n-1) x 2
Therefore, n = (200 + 2 )/2 = 101
That is, there are 101 even integers between 99 and 301.
Sum of the 'n' terms in an AP = n/2 ( first term + last term)
= 101/2 (300+100)
= 20200
Thus sum of all the even integers between 99 and 301 = 20200
Learn more about arithmetic progressions at brainly.com/question/24592110
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Answer:
I think it is c sorry if it's worng
Step-by-step explanation:
8(2x - 3)-6x
First, replace all the x's with 3s. Simplfy 2 × 3 to 6. / Your problem should look like: 8 × (6 - 3) - 6 × 3
Second, simplify 6 - 3 to 3. / Your problem should look like: 8 × 3 - 6 × 3
Third, simplify 8 × 3 to 24. / Your problem should look like: 24 - 6 × 3
Fourth, simplify 6 × 3 to 18. / Your problem should look like: 24 - 18
Fifth, simplify 24 - 18 to get 6. / Your problem should look like: 6
Answer: 6
Answer:
32
Step-by-step explanation:
math
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