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AleksAgata [21]
3 years ago
7

Can someone find the area for this shape in please quick

Mathematics
2 answers:
Juli2301 [7.4K]3 years ago
8 0

Answer:

im pretty sure the answer would be 40m

Step-by-step explanation:

Cerrena [4.2K]3 years ago
6 0

Answer:

Area: 40m²

Step-by-step explanation:

The first thing we need to do in order to solve this problem is split the shape into three different rectangles. We are going to have one rectangle that is 3 m by 7 m, one that's 4 by 4, and one that is 3 by 1. We are now going to find the area of each one individually.

To find the area of a rectangle, you multiply the length by the width. So, for the first rectangle, you would multiply 7 and 3. This equals 21, so that is the area of the first rectangle. Repeat this for the other two rectangles, then add all the areas together and you will get the total area of this shape.

<em>Remember, when you are finding the area of an object, you must </em><em>always </em><em>put the </em><u><em>label </em></u><em>and a </em><u><em>squared sign</em></u><em> next to the number!</em>

7×3=21

4×4=16

3×1=3

21+16+3=40

Area = 40m²

I hope this is fairly easy to understand. Please let me know if you need any more help. Good luck and have a great morning!

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What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10
jasenka [17]

<u>Answer:</u>

 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

<u>Solution: </u>

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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