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elixir [45]
3 years ago
6

(ii) 4xz? [- xyz + 3yz?]​

Mathematics
1 answer:
Goshia [24]3 years ago
3 0

Answer: -4x²z²y + 12xz²y

Step-by-step explanation:-------------

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Find the series in A.P whose sum to n terms is n*(n+2)​
Lemur [1.5K]

Answer:

3, 5, 7, 9, 11, .........

Step-by-step explanation:

Given

S_{n} = n(n + 2) , then

S₁ = 1(1 + 2) = 1(3) = 3 ⇒ a₁ = 3

S₂ = 2(2 + 2) = 2(4) = 8

S₃ = 3(3 + 2) = 3(5) = 15

Thus

a₂ = S₂ - S₁ = 8 - 3 = 5

a₃ = S₃ - S₂ = 15 - 8 = 7

The first 3 terms are 3, 5, 7

This is an AP with common difference d = 2, then

a₄ = a₃ + 2 = 7 + 2 = 9

a₅ = a₄ + 2 = 9 + 2 = 11

and so on

3 0
3 years ago
16 increased by twice a number is -24
wlad13 [49]
X-the number

16 + 2x = -24    |subtract 16 from both sides

2x = -40      |divide both sides by 2

<u>x = -20</u>

Answer: <em>The number is -24</em>
8 0
3 years ago
What is 126/3 simplafiyed
spayn [35]

Answer:

42

Step-by-step explanation:

42 is the answer

5 0
3 years ago
Read 2 more answers
Santiago walked 500 m
N76 [4]

Answer:

500 X 7 = 3500, 3500m = 3.5 km

She walked 3.5 km in 1 week.

7 0
3 years ago
Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans
zloy xaker [14]
Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.



Part B:

The area of the square is given by

Area=(x+4)^2=x^2+8x+16

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

x^2+8x+16=6x+8 \\  \\ \Rightarrow x^2+8x-6x+16-8=0 \\  \\ \Rightarrow x^2+2x+8=0 \\  \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2}  \\  \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2}  \\  \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
7 0
3 years ago
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