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3 years ago

Help please!!

2 answers:

6 0

The perimeter of a rectangle is 168 ft. The ratio of the length to the width is 11:1. You are asked to find the width of the rectangle.
L/W = 11
L = 11W
P = 2L + 2W
168 = 2(11W) + 2W
168 = 24W
W = 7 ft

The answer is letter A.

balu736 [363]3 years ago

3 0

12 pieces in total so (168/2):12=7
the width is either 7, either 7*11=77

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2 years ago

Read 2 more answers

What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10

jasenka [17]

Answer:

The coefficient of $x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552$

Solution:

The given expression is $(2 x-3 y)^{10}$

As per binomial theorem, we know,

$(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}$

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now $x^{5}\times y^5$ will be the 6 term where k =5

Now, $\mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}$

So, the coefficient of $x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right$ .

$10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252$

The coefficient of $x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552$

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3 years ago