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blsea [12.9K]
3 years ago
10

Please help me with this question ​

Mathematics
1 answer:
umka21 [38]3 years ago
8 0

Answer:

-3x^2+16x+64

Step-by-step explanation:

First, you need to find the area of both areas (the total area and the area of the unshaded part). Let's find the total area.

(x+8)(x+8)=x^2+16x+64

The total area is x^2+16x+64

(2x)(2x)=4x^2

The area of the unshaded part is 4x^2

Now we have to subtract to find the area of the shaded part.

x^2+16x+64-4x^2

-3x^2+16x+64

That is the answer

Hope this helped!! :D

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(show the supposition, proof and conclusion)
larisa86 [58]

Answer:

Step-by-step explanation:

We are given that a and b are rational numbers where b\neq0 and x is irrational number .

We have to prove a+bx is irrational number by contradiction.

Supposition:let  a+bx is a rational number then it can be written in \frac{p}{q} form

a+bx=\frac{p}{q} where q\neq0 where p and q are integers.

Proof:a+bx=\frac{p}{q}

After dividing p and q by common factor except 1 then we get

a+bx=\frac{r}{s}

r and s are coprime therefore, there is no common factor of r and s except 1.

a+bx=\frac{r}{s} where r and s are integers.

bx=\frac{r}{s}-a

x=\frac{\frac{r}{s}-a}{b}

When we subtract one rational from other rational number then we get again a rational number and we divide one rational by other rational number then we get quotient number which is also rational.

Therefore, the number on the right hand of equal to is rational number but x is a irrational number .A rational number is not equal to an irrational number .Therefore, it is contradict by taking a+bx is a rational number .Hence, a+bx is an irrational number.

Conclusion: a+bx is an irrational number.

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3 years ago
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