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3 years ago

Which of the following expressions is equivalent to -6?

Mathematics

1 answer:

djyliett [7]3 years ago

7 0

Answer:

-12 + 6

Step-by-step explanation:

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Amanda exercised for 10 minutes every day in the first week, 20 minutes in the second week, 30 minutes in the third week, and 40

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Amanda Billy
1st week 10 5
2nd week 20 10
3rd week 30 20
4th week 40 40
<span>
A) Amanda's method is linear because the number of minutes increased by an equal number every week.</span>
common difference is 10.
1st week 0 + 10 = 10
2nd week 10 + 10 = 20
3rd week 20 + 10 = 30
4th week 30 + 10 = 40

Billy's method is exponential:
5(2)^x

1st week 5(2⁰) = 5(1) = 5
2nd week 5(2¹) = 5(2) = 10
3rd week 5(2²) = 5(4) = 20
4th week 5(2³) = 5(8) = 40

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A gardener expects that 75% of the seeds she plants will produce plants she wants 45 plants how many seeds should she plant ?

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Let it be x
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3 years ago

What are the solutions of the equation 9x4 – 2x2 – 7 = 0? Use u substitution to solve. tions of the equation 9

skelet666 [1.2K]

Answer:

Step-by-step explanation:

Let $u^2=x^4\\u = x^2$

Subbing in:

$9u^2-2u-7=0$

a = 9, b = -2, c = -7

The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:

$9u^2-9u+7u-7=0$

Group together in groups of 2:

$(9u^2-9u)+(7u-7)=0$

Now factor out what's common within each set of parenthesis:

$9u(u-1)+7(u-1)=0$

We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:

$(u-1)(9u+7)=0$

Remember that $u=x^2$

so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.

$(x^2-1)(9x^2+7)=0$

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring $(x^2-1)$ gives us that x = 1 and -1. The other set is a bit more tricky. If