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3 years ago

7. Factor the expression. 42 - 64

Mathematics

1 answer:

Dmitriy789 [7]3 years ago

3 0

Answer:

-22 = -2^1×11^1

Step-by-step explanation:

Factor the following integer:

-22

Hint: | Is 22 divisible by 2?

The last digit of 22 is 2, which means it is even. Therefore 22 is divisible by 2:

22 = 2 11:

-22 = -2×11

Hint: | Now try to factor 11. Is 11 divisible by 2?

11 is not divisible by 2 since 11 is odd and 2 is even:

-22 = -2×11 (11 is not divisible by 2)

Hint: | Is 11 divisible by 3?

The sum of the digits of 11 is 1 + 1 = 2, which is not divisible by 3. This means 11 is not divisible by 3:

-22 = -2×11 (11 is not divisible by 2 or 3)

Hint: | Is 11 divisible by 5?

The last digit of 11 is not 5 or 0, which means 11 is not divisible by 5:

-22 = -2×11 (11 is not divisible by 2, 3 or 5)

Hint: | Is 11 divisible by 7?

Divide 7 into 11:

| | 1 | (quotient)

7 | 1 | 1 |

- | | 7 |

| | 4 | (remainder)

11 is not divisible by 7:

-22 = -2×11 (11 is not divisible by 2, 3, 5 or 7)

Hint: | Is 11 prime?

No primes less than 11 divide into it. Therefore 11 is prime:

-22 = -2×11

Hint: | Express -22 as a product of prime powers.

There is 1 copy of 2 and 1 copy of 11 in the product:

Answer: -22 = -2^1×11^1

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Step-by-step explanation:

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3 years ago

If a(x) = 3x + 1 and b(x)= squareroot x-4, what is the domain of (b*a)(x)?

oksano4ka [1.4K]

$\boxed{ \ x \geq 1 \ }$ or can be written as $\boxed{ \ [1, \infty) \ }$

<h3>Further explanation</h3>

This is a question about the composition of functions and how to get a domain function.

Given $\boxed{ \ a(x) = 3x + 1 \ }$ and $\boxed{ \ b(x) = \sqrt{x - 4} \ }$ .

We will form (b o a)(x) and then determine the domain.

$\boxed{ \ (b \circ a)(x) = b(a(x)) \ }$

Replace each appearance of x in b(x) with $\boxed{ \ a(x) = 3x + 1 \ }$ .

$\boxed{ \ (b \circ a)(x) = \sqrt{(3x + 1) - 4} \ }$

Thus, $\boxed{ \ (b \circ a)(x) = \sqrt{3x - 3} \ }$

To be defined, the value under the radical sign must not be negative. Therefore, the domain of $(b \circ a)(x) = \sqrt{3x - 3}$ are processed as follows.

$\boxed{ \ 3x - 3 \geq 0 \ }$

Both sides added by 3.

$\boxed{ \ 3x \geq 3 \ }$

Both sides divided by 3.

$\boxed{ \ x\geq 1 \ }$

Thus, the domain of $(b \circ a)(x) = \sqrt{3x - 3}$ is $\boxed{ \ x \geq 1 \ }$ or can be written as $\boxed{ \ [1, \infty) \ }$

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