What is Decrease 90 by 1/2

Mathematics

2 answers:

devlian [24]3 years ago

8 0

Answer:

Step-by-step explanation:

$\frac{1}{2}$ × 90 = 45

Decrease means to subtract the amount by 45, then

90 - 45 = 45

svetlana [45]3 years ago

6 0

The answer is 45. You just divide 90 by 2 or times 90 by half.

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In which expression is the number 5 a factor? A. z+5 B. z-5 C.5z D.5/z

LenaWriter [7]

The correct answer is: C.5z

Further explanation:

A factor is a number that is multiplied with a number

Let us look at the options one by one

A. z+5

5 is not a factor as it is being added to z.

B. z-5

5 is not a factor as it is being subtracted from z.

C.5z

5 is a factor as it is being multiplied with the z and can be separated easily

Hence,

The correct answer is: C.5z

Keywords: Factors, Multiple Choice

Learn more about factors at:

#LearnwithBrainly

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3 years ago

What transformation will the following equation have from its parent?

muminat

Reflection across the x-axis (because there is a negative)

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2 years ago

Factor completely x^8- 16.

Nataly_w [17]

Answer:

Step-by-step explanation:

So in this example we'll be using the difference of squares which essentially states that: $(a-b)(a+b)=a^2-b^2$ or another way to think of it would be: $a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$ . So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:

$x^8-16 = (x^4-4)(x^4+4)$

Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:

$x^4-4 = (x^2-2)(x^2+2)$

So completely factored form is: $(x^2-2)(x^2+4)(x^4+4)$

I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because $x^2+4 = x^2-(-4)$ . and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: $x^4+4 = x^4 - (-4) = (x^2-\sqrt{-4})(x^2+\sqrt{-4}) = (x^2-2i)(x^2+2i)$ just something that might be useful in some cases.