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2 years ago

Please help! I need answers for both questions it’s very confusing!

Mathematics

2 answers:

SpyIntel [72]2 years ago

6 0

When it seems the negative number get bigger, they’re actually getting smaller. For example -13 is greater than -20. Plus, the absolute value would always be positive.

The answer to number 1 would be:

-15 < -8 8/9 < -3.3

The answer to number 2 would be:

15 < -3.3 < -8 8/9

I hope this helped, and have an awesome day! <3

Lilit [14]2 years ago

5 0

Answer:

for number 5 its:

-15 -8 8/9 -3.3

and for number 6 its:

-3.3 -8 8/9 and -15

also sorry if its wrong

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Ten large marbles weigh as much as 14 small marbles. How many large marbles weigh as much as 35 small marbles?

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10 large ------------- 14 small
x large -------------- 35 small
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10*35=x*14
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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]

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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