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3 years ago

Four different ' four 4's problems ( where they all equal a different #) and four different four 4's that equal the same #

Mathematics

1 answer:

saveliy_v [14]3 years ago

5 0

Answer:

1=(4/4)*(4/4)

2=(4/4)+(4/4)

3 = (4+4+4)/4

4 = 4 + (4-4)*4

20=4 × (4/4 + 4)

20=4 × (4/4 - (-4))

$20 = \sqrt{4} +\sqrt{4} + (4x4)$

$20 = \sqrt{4} +\sqrt{4} + (-4 x -4)$

Step-by-step explanation:

The x is a multiplication sign not a variable, I hope this right. Good luck, I am so tired.

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Answer:

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Step-by-step explanation:

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2 years ago

What is the fourth term in the binomial expansion of (2x – y) 7?

nata0808 [166]

Answer:

$(7C4) (2x)^4 (-y)^{7-4}$

And replacing we got:

$35 (2^4) x^4 (-y)^{-3}$

And then the final term would be:

$-560 x^4 y^3$

Step-by-step explanation:

For this case we have the following expression:

$(2x-y)^7$

And we can use the binomial theorem given by:

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And for this case we want to find the fourth term and using the formula we have:

$(7C4) (2x)^4 (-y)^{7-4}$

And replacing we got:

$35 (2^4) x^4 (-y)^{-3}$

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Read 2 more answers

Find the values of k so that each remainder is three. 10. (x^2+ 5x + 7) = (x + k)

Goryan [66]

Answer:

$k=1\text{ or } k=4$

Step-by-step explanation:

We can use the Polynomial Remainder Theorem. It states that if we divide a polynomial P(x) by a binomial in the form (x - a), then our remainder will be P(a).

We are dividing:

$(x^2+5x+7)\div(x+k)$

So, a polynomial by a binomial factor.

Our factor is (x + k) or (x - (-k)). Using the form (x - a), our a = -k.

We want our remainder to be 3. So, P(a)=P(-k)=3.

Therefore:

$(-k)^2+5(-k)+7=3$

Simplify:

$k^2-5k+7=3$

Solve for k. Subtract 3 from both sides:

$k^2-5k+4=0$

Factor:

$(k-1)(k-4)=0$

Zero Product Property:

$k-1=0\text{ or } k-4=0$

Solve:

$k=1\text{ or } k=4$

So, either of the two expressions:

$(x^2+5x+7)\div(x+1)\text{ or } (x^2+5x+7)\div(x+4)$

Will yield 3 as the remainder.

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3 years ago

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Answer:

Step-by-step explanation:

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