Crediting an account with a gift card of $60 positive or negative $60

Mathematics

1 answer:

Marina CMI [18]3 years ago

4 0

Postive because if your adding something to it is now postive

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Binomial expansion of (x+2)^4

N76 [4]

The two terms are x and 2, thus, x+2 is a binomial. We have to multiply the binomial by itself four times since it is raised to 4th power.

Let us multiply x+2 by itself using Polynomial Multiplication:

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

Taking the result, let us multiply it again by a+b:

(x^2 + 4x + 4)(x+2) = x^3 + 6x^2 + 12x + 8

And again:

(x^3 + 6x^2 + 12x + 8)(x+2) = x^4 + 8x^3 + 24x^2 + 32x +16

The binomial expansion of (x+2)^4 is x^4 + 8x^3 + 24x^2 + 32x +16

4 0

3 years ago

The sum of the digits is 11. When rounded to the nearest hundred, it's 500. Rounding to the nearest 10 makes it 530. What's the

kupik [55]

If rounded to the nearest 10 is 530, then this number must be between 525 and 534. The only one which digits sum up to 11 is 533.

4 0

4 years ago

Read 2 more answers

What is the value of the expression if m = –5 and n = 3? Negative StartFraction 24 Over 25 EndFraction Negative StartFraction 4

KatRina [158]

You can use the fact that a number raised to negative powers can be written as 1 divided by that number raised to positive number.

The value of the given expression when m = -5 and n= 3 is $-\dfrac{2}{5}$

<h3>How can we rewrite a number raised to negative power?</h3>

Suppose we've got $a^{-b}$ . then it can be rewritten as

$a^{-b} = \dfrac{1}{a^b}$

It is because $a^0 = 1$ for any a except 0, and that $\dfrac{a^0}{a^b} = a^{0-b} = a^{-b}$

If base are same and there is multiplication, then there will be addition in power.

Also, know the fact that if base are same and there is division, there will be subtraction in power(subtracting since that denominator when goes up, its power gets negated(gets negative sign to the existing power)

Thus,

$a^b \timesa ^c = a^{b+c}\\\\\dfrac{a^b}{a^c} = a^{b-c}$

<h3>Using above method to simplify the given expression before evaluating</h3>

The given expression is $\dfrac{2m^{-1}n^5}{3m^0n^4} = \dfrac{2n^5}{3m^0m^1n^4} = \dfrac{2n^5}{3mn^4} = \dfrac{2n^{5-4}}{3m} = \dfrac{2n^1}{3m} = \dfrac{2n}{3m}$