P = 2(L + W)
L = W + 5
A = 4P + 2
P = 2(W + 5 + W)
P = 2(2W + 5)
P = 4W + 10
A = 4P + 2
A = 4(4W + 10) + 2
A = 16W + 42
A = L * W
A = W(W + 5)
A = W^2 + 5W
W^2 + 5W = 16W + 42
W^2 + 5W - 16W - 42 = 0
W^2 - 11W - 42 = 0
(W + 3)(W - 14) = 0
W - 14 = 0
W = 14 <==
L = W + 5
L = 14 + 5
L = 19 <==
P = 2(19 + 14)
P = 2(33)
P = 66
A = L * W
A = 19 * 14
A = 266
answer : length = 19, width = 14....perimeter = 66....area = 266
Answer:
Yes
Step-by-step explanation:
Given that in the June 2007 issue, Consumer Reports also examined the relative merits of top-loading and front-loading washing machines, testing samples of several different brands of each type.
The difference in mean values test gave a p value of 0.32
Confidence level = 95%
Alpha = 1-0.95 = 0.05
Compare p with alpha, here p >alpha
Hence we accept null hypothesis that there is no difference in the means.
Confidence interval method also will yield the same result. i.e. confidence interval for difference of means would definitely contain 0 at 95% conf level.
So answer is yes
U/9 = 8/12 u = 6
Step 1: Cancel the common factor (4)
u = 2
—- —-
9 3
Step 2: multiply both sides by 9
9u 2 * 9
—- = ——-
9 3
Step 3: simplify
2 *9 = 18
18 ÷ 3 = 6
u = 6
You just want to simplify right?!
45. (a^2b^3)(ab)^-2
= (a^2b^3)(a^-2b^-2)
= b
46. (-3x^3y)^2(4xy^2)
= (-9x^6y^2)(4xy^2)
= -36x^7y^4
47. 3c^2d(2c^3d^5) / 15c^4d^2
= 6c^5d^6 / 15c^4d^2
= 2/5c1/4x^4
48. -10g^6h^9(g^2h^3) / 30g^3h^3
= -10g^8h^12 / 30g^3h^3
= -1/3g^5h^9
49. 5x^4y^2(2x^5y^6) / 20x^3y^5
= 10x^9y^8 / 20x^3y^5
= 1/2x^6 1/3y^3
50. -12n^7p^5(n^2p^4) / 36n^6p^7
= -12n^9p^9 / 36n^6p^7
= -1/3n^3p^2
(Sorry it’s messy it’d look better if my phone could actually put the numbers to the power)
