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2 years ago

In the equation y=-x 4 what is the unit rate?

Mathematics

1 answer:

ioda2 years ago

4 0

The unit rate is -1. because the unit rate has to have a denominator of 1.

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Is 11/20 greater than a half

Rainbow [258]

Yes... a half is equal to 10/20 and 11/20 is greater than 10/20 so 11/20 is greater than a half

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2 years ago

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Matthew and Lizbeth are college students. Matthew has $87 in his bank account. He spends $45 on groceries. He deposits $30 into

Murljashka [212]

Answer:

What is the question to this

Step-by-step explanation:

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2 years ago

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4t2–40t+84 factorise quadratics

Evgen [1.6K]

Answer:

4( t - 7) ( t - 3 )

Step-by-step explanation:

4t² - 40t + 84

4(t² - 10t + 21)

4( t - 7) ( t - 3 )

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Find the equivalent exponential expression. [(-4)^5]^3

Sunny_sXe [5.5K]

Answer:

$-4^{15}$

Step-by-step explanation:

Given expression:

$\left[(-4)^5\right]^3$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\implies (-4)^{(5 \times 3)}$

$\implies (-4)^{15}$

$\textsf{Apply exponent rule} \quad (-a)^n=-a^n,\:\: \textsf{ if }n \textsf{ is odd}:$

$\implies -4^{15}$

4 0

1 year ago

How many different ways are there to choose a subset of the set {1,2,3,4,5,6} so that the product of the members of the subset i

zvonat [6]

You have to pick at least one even factor from the set to make an even product.

There are 3 even numbers to choose from, and we can pick up to 3 additional odd numbers.

For example, if we pick out 1 even number and 2 odd numbers, this can be done in

$\dbinom 31 \dbinom 32 = 3\cdot3 = 9$

ways. If we pick out 3 even numbers and 0 odd numbers, this can be done in

$\dbinom 33 \dbinom 30 = 1\cdot1 = 1$

way.

The total count is then the sum of all possible selections with at least 1 even number and between 0 and 3 odd numbers.

$\displaystyle \sum_{e=1}^3 \binom 3e \sum_{o=0}^3 \binom 3o = 2^3 \sum_{e=1}^3 \binom3e = 8 \left(\sum_{e=0}^3 \binom3e - \binom30\right) = 8(2^3 - 1) = \boxed{56}$

where we use the binomial identity

$\displaystyle \sum_{k=0}^n \binom nk = \sum_{k=0}^n \binom nk 1^{n-k} 1^k = (1+1)^n = 2^n$

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11 months ago