This is an arithmetic sequence with the common difference being 1
and if u have 1 cent on day 1 and 2 cents on day 2....then on day 40, u would have 40 cents....but we need the sum of all the days..
the sum of an arithmetic sequence can be found using the formula :
sn = n(a1 + an) / 2
n = number of terms = 40
a1 = first term = 1
an = last term = 40
now we sub
s(40) = 40(1 + 40) / 2
s(40) = 40(41)/2
s(40) = 1640/2
s(40) = 820.....820 cents = $ 8.20 <== the sum
Answer The unit rate in calories per cracker is 7.
(I'll add explanation if needed)
I hope this helped! :)
(2x-3y)^5
(2x-3y)(2x-3y)(2x-3y)(2x-3y)(2x-3y)
1st and 2nd power :
(2x-3y)(2x-3y) = 2x(2x-3y)-3y(2x-3y) = 4x² - 6xy - 6xy + 9y²
= 4x² - 12xy + 9y²
3rd power:
(2x-3y)(4x² - 12xy + 9y²) = 2x(4x² - 12xy + 9y²) - 3y(4x² - 12xy + 9y²)
8x³ - 24x²y + 18xy² - 12x²y +36xy² - 27y³
8x³ - 24x²y - 12x²y + 18xy² + 36xy² - 27y³
8x³ - 36x²y + 54xy² - 27y³
4th power
(2x-3y)(8x³ - 36x²y + 54xy² - 27y³) = 2x(8x³ - 36x²y + 54xy² - 27y³) -3y(8x³ - 36x²y + 54xy² - 27y³) = 16x^4 - 72x³y + 108x²y² - 54xy³ - 24x³y + 108x²y² - 162xy³ + 81y^4
16x^4 - 72x³y - 24x³y + 108x²y² + 108x²y² - 54xy³ - 162xy³ + 81y^4
16x^4 - 96x³y + 216x²y² - 216xy³ + 81y^4
5th power
(2x-3y)(<span>16x^4 - 96x³y + 216x²y² - 216xy³ + 81y^4)
2x(</span>16x^4 - 96x³y + 216x²y² - 216xy³ + 81y^4) - 3y(<span>16x^4 - 96x³y + 216x²y² - 216xy³ + 81y^4)
= 32x^5 - 192x^4y + 432x</span>³y² - 432x²y³ + 162xy^4 - 48x^4y + 288x³y² - 648x²y³ + 648xy^4 - 243y^5
32x^5 - 192x^4y -48x^4y + 432x³y² + 288x³y² - 432x²y³ - 648x²y³ + 162xy^4 + 648xy^4 - 243y^5
32x^5 - 240x^4y + 720x³y² - 1,080x²y³ + 810xy^4 - 243y^5
Closest one is 4.9cm
✖️multiply
3.141592654....as in in pi with the
radius of 1.25^2to the power of 2
or 1.562 times pi
Answer:
9
Step-by-step explanation:
Given: Subtract 9 from 10. Multiply this difference by 3, square this product.
Now, lets simplify it.
![[(10-9)\times 3]^{2}](https://tex.z-dn.net/?f=%5B%2810-9%29%5Ctimes%203%5D%5E%7B2%7D)
Using PEDMAS rule to solve it
First working on Parenthesis
= ![[(1)\times 3]^{2}](https://tex.z-dn.net/?f=%5B%281%29%5Ctimes%203%5D%5E%7B2%7D)
Opening the parenthesis
= ![[1\times 3]^{2}](https://tex.z-dn.net/?f=%5B1%5Ctimes%203%5D%5E%7B2%7D)
= ![[3]^{2}](https://tex.z-dn.net/?f=%5B3%5D%5E%7B2%7D)
= 
∴ Answer is 9.
