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Morgarella [4.7K]
2 years ago
15

From the sum of 4a + b + c and 9a - c subtract the sum of 18a - b - c and 14a - 2b - c

Mathematics
2 answers:
sergiy2304 [10]2 years ago
5 0

1:-

\\ \rm\Rrightarrow 4a+b+c+9a-c

\\ \rm\Rrightarrow 4a+9a+b+c-c

\\ \rm\Rrightarrow 13a+b

2:-

\\ \rm\Rrightarrow 18a-b-c-(14a-2b-c)

\\ \rm\Rrightarrow 18a-b-c-14a+2b+c

\\ \rm\Rrightarrow 18a-14a-b+2b-c+c

\\ \rm\Rrightarrow 4a+b

cricket20 [7]2 years ago
3 0

Answer:

-19c + 4b + 2c

Step-by-step explanation:

(4a + b + c + 9a - c) - (18a - b - c + 14a - 2b - c)

= (13a + b) - (32a - 3b - 2c)

= 13a + b - 32a + 3b + 2c

= -19c + 4b + 2c

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What are the values of X and Z.
stiv31 [10]

Answer:

x = 12, z  = 86

Step-by-step explanation:

Two angles are called supplementary when their measures add up to 180 degrees.

(12x -58  + 13x - 62 ) = 180°

12 x + 13x -58 - 62 = 180°

25x - 120 = 180

25x = 180 + 120

x = 300/ 25

x = 12

----------------------------------------------------------------------------------

(z° + 13x - 62) = 180

(z° + 13(12) - 62 )= 180

(z° +  156 - 62 )= 180

(z° + 94 ) = 180

z° = 180 - 94

z  = 86

5 0
3 years ago
A 25 gram sample of a sample that's used for drug research has a k value of 0.1205
ra1l [238]
N=12.5(at time t=t) Here half life
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Your answer should be 5.75225876
4 0
3 years ago
If a cube has a<br> volume of 8 cm<br> what is it's side<br> length?
lana [24]
Side length: 2 cm.

Volume of a cube:
V=a^3
(where V is volume, and a is side length)

If v=8, the formula would now be 8=a^3

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If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.
Lelechka [254]

Answer:

\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)

So we can say that

a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab

and we can notice that

a^2+b^2=(a+b)^2-2ab so

(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab

it comes

x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}

multiply by 3

3x^2-20x+2*16+5=3x^2-20x+37

4 0
3 years ago
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