You're not dumb! Don't say that :)
Answer: A. Yes
B. 1 feet by 11 feet
Step-by-step explanation: Area = length x width
Answer:
(18x)-3
Step-by-step explanation:
Given: Yu, Nailah, and Elena each bought between 7 and 9 yards of ribbon Yu bought 3 pieces of ribbon. Nailah bought 5 pieces of ribbon. Elena bought 6 pieces of ribbon.
To find: Who can buy which ribbon
Solution:
Ribbon Sizes
1 2/3 = 5/3 yard
4/5 yard
3 1/2 = 7/2 Yard
Yu, Nailah, and Elena each bought between 7 and 9 yards of ribbon
Yu bought 3 pieces of ribbon
=> 3 * 5/3 = 5
3 * 4/5 = 2.4
3 * 7/2 = 10.5
Nailah bought 5 pieces of ribbon
=> 5 * 5/3 = 8.33
5 * 4/5 = 4
5 * 7 /2 = 17.5
Elena bought 6 pieces of ribbon
=> 6 * 5/3 = 10
6 * 4/5 = 4.8
6 * 7 /2 = 21
Only value between 7 & 9 is 5 * 5/3 = 8.33
hence Nailah only can buy 1 2/3 = 5/3 yard ribbon
or there is some mistake in the data

so, as you can see above, the common ratio r = 1/2, now, what term is +4 anyway?


so is the 8th term, then, let's find the Sum of the first 8 terms.

![\bf S_8=512\left[ \cfrac{1-\left( \frac{1}{2} \right)^8}{1-\frac{1}{2}} \right]\implies S_8=512\left(\cfrac{1-\frac{1}{256}}{\frac{1}{2}} \right)\implies S_8=512\left(\cfrac{\frac{255}{256}}{\frac{1}{2}} \right)\\\\\\S_8=512\cdot \cfrac{255}{128}\implies S_8=1020](https://tex.z-dn.net/?f=%20%5Cbf%20S_8%3D512%5Cleft%5B%20%5Ccfrac%7B1-%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E8%7D%7B1-%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cright%5D%5Cimplies%20S_8%3D512%5Cleft%28%5Ccfrac%7B1-%5Cfrac%7B1%7D%7B256%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%20%5Cright%29%5Cimplies%20S_8%3D512%5Cleft%28%5Ccfrac%7B%5Cfrac%7B255%7D%7B256%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%20%5Cright%29%5C%5C%5C%5C%5C%5CS_8%3D512%5Ccdot%20%5Ccfrac%7B255%7D%7B128%7D%5Cimplies%20S_8%3D1020%20)
Important: Use the symbol "^" to denote exponentiation:
<span>x3 – 9x2 + 5x – 45 NO
</span><span>x^3 – 9x^2 + 5x – 45 YES
Look at the first 2 terms. They can be rewritten as x^2(x-9). Then look at the last 2 terms. They can be rewritten as 5(x-9). So, x-9 is the common factor here. Thus, the original expression becomes:
(x^2-5)(x-9).
Note that x^2-5 can be factored, so that the final 3 factors are:
(x-sqrt(5)), (x+sqrt(5)), (x-9).</span>