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meriva
3 years ago
9

Can someone help me with this?

Mathematics
1 answer:
Alla [95]3 years ago
5 0

Answer:

d and e

Step-by-step explanation:

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STILL GOING BAN ON THEM ANYWAY
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Answer:

I-

Step-by-step explanation:

Um okay.

7 0
2 years ago
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You have an equation y=x+7. Is the ordered pair (1,6) a solution? Yes or no?
vaieri [72.5K]
Answer:
no

step-by-step explanation:
y=x+7
x is the slope of 1
7 is the y-intercept
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8 0
2 years ago
The first modern humans appeared on Earth about 200,000 years ago. Would it be possible to measure that number in terms of feet
Marina86 [1]
Hi um i know this answer is 79
8 0
2 years ago
HELPP!!!!!!!!!!!!!!!!
Anvisha [2.4K]

Answer:

Step-by-step explanation:

We'll take this step by step.  The equation is

8-3\sqrt[5]{x^3}=-7

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time.  First thing is to subtract 8 from both sides:

-3\sqrt[5]{x^3}=-15

The goal is to isolate the term with the x in it, so that means that the -3 has to go.  Divide it away on both sides:

\sqrt[5]{x^3}=5

Let's rewrite that radical into exponential form:

x^{\frac{3}{5}}=5

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely.  On the right, let's rewrite that back in radical form to solve it easier:

x=\sqrt[3]{5^5}

Let's group that radicad into groups of 3's now to make the simplifying easier:

x=\sqrt[3]{5^3*5^2} because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:

x=5\sqrt[3]{5^2} which is the same as:

x=5\sqrt[3]{25}

8 0
2 years ago
Anyone want to ch at?
algol13

Answer:

sure?

Step-by-step explanation:

7 0
2 years ago
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