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alexgriva [62]
2 years ago
11

Solve using elimination. –8x + 6y = –16 –8x + 9y = 8

Mathematics
1 answer:
stira [4]2 years ago
5 0
I think this is right

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There are 4 doughnuts in 1/3 of a dozen
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In London today, four times the high temperature was more than twice the high temperature plus
4vir4ik [10]

Answer:

Let's define the high temperature as T.

We know that:

"four times T, was more than 2*T plus 66°C"

(i assume that the temperature is in °C)

We can write this inequality as:

4*T > 2*T + 66°C

Now we just need to solve this for T.

subtracting 2*T in both sides, we get:

4*T - 2*T > 2*T + 66°C - 2*T

2*T > 66°C

Now we can divide both sides by 2:

2*T/2 > 66°C/2

T > 33°C

So T was larger than 33°C

Notice that T = 33°C is not a solution of the inequality, then we should use the symbol ( for the set notation.

Then the range of possible temperatures is:

(33°C, ...)

Where we do not have an upper limit, so we could write this as:

(33°C, ∞°C)

(ignoring the fact that ∞°C is something impossible because it means infinite energy, but for the given problem it works)

4 0
2 years ago
Try this hard Math Problem if you dare!!
Dennis_Churaev [7]

Answer:

a. (x - 3)^2 + 16

b. 8(x -7)^2

c. (a^2 - 1)(7x - 6) or (a+1)(a-1)(7x-6)

d. (x^2-4)(x^2+3) or (x-2)(x+2)(x^2+3)

e. (a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})

Step-by-step explanation:

a.\ (x + 1)^2 - 8(x - 1) + 16

Expand

(x + 1)(x + 1) - 8(x - 1) + 16

Open brackets

x^2 + x + x + 1 - 8x + 8 + 16

x^2 + 2x + 1 - 8x + 24

Collect Like Terms

x^2 + 2x - 8x+ 1  + 24

x^2 - 6x+ 25

Express 25 as 9 + 16

x^2 - 6x+ 9 + 16

Factorize:

x^2 - 3x - 3x + 9 + 16

x(x -3)-3(x - 3) + 16

(x - 3)(x - 3) + 16

(x - 3)^2 + 16

b.\ 8(x - 3)^2 - 64(x-3) + 128

Expand

8(x - 3)(x - 3) - 64(x-3) + 128

8(x^2 - 6x+ 9) - 64(x-3) + 128

Open Brackets

8x^2 - 48x+ 72 - 64x+192 + 128

Collect Like Terms

8x^2 - 48x - 64x+192 + 128+ 72

8x^2 -112x+392

Factorize

8(x^2 -14x+49)

Expand the expression in bracket

8(x^2 -7x-7x+49)

Factorize:

8(x(x -7)-7(x-7))

8((x -7)(x-7))

8(x -7)^2

c.\ 7a^2x - 6a^2 - 7x + 6

Factorize

a^2(7x - 6) -1( 7x - 6)

(a^2 - 1)(7x - 6)

The answer can be in this form of further expanded as follows:

(a^2 - 1^2)(7x - 6)

Apply difference of two squares

(a+1)(a-1)(7x-6)

d.\ x^4 - x^2 - 12

Express x^4 as x^2

(x^2)^2 - x^2 - 12

Expand

(x^2)^2 +3x^2- 4x^2 - 12

x^2(x^2+3) -4(x^2+3)

(x^2-4)(x^2+3)

The answer can be in this form of further expanded as follows:

(x^2-2^2)(x^2+3)

Apply difference of two squares

(x-2)(x+2)(x^2+3)

e.\ a^{4n} -b^{4n}

Represent as squares

(a^{2n})^2 -(b^{2n})^2

Apply difference of two squares

(a^{2n} -b^{2n})(a^{2n} +b^{2n})

Represent as squares

((a^{n})^2 -(b^{n})^2)(a^{2n} +b^{2n})

Apply difference of two squares

(a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})

6 0
3 years ago
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