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Ira Lisetskai [31]
4 years ago
14

What is the GCF of 30b^2, 80b^2, and 20b^3?

Mathematics
1 answer:
Nana76 [90]4 years ago
5 0
30b^2, 80b^2, \ and \ 20b^3

30b^2 = 10b^2(3)

80b^2 = 10b^2(8)

20b^3 = 10b^2(2b)

\text {GCF = } 10b^2



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Answer:

\vec{r(t)}=t\hat{i}+6t^2\hat{j}+(3t^2+72t^4)\hat{k} \text{ For }t \in (-\infty, \infty)

Step-by-step explanation:

First plug the equation of y into the equation of z, so that we get their intersection:

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Into: z=3x^2+2y^2

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Then we set  

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And plug that in the equation y=6x^2 and the one we just got for z.

So that we get:

y=6t^2, z=3t^2+72t^4

Therefore, the vector function that represents the curve of intersection is:

\vec{r(t)}=t\hat{i}+6t^2\hat{j}+(3t^2+72t^4)\hat{k} \text{ For }t \in (-\infty, \infty)

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<h3>What absolute value equation can be written by Edwin to find the numbers which are 0.3 units away from 3.9 on the number line?</h3>

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Read more on absolute value equation;

brainly.com/question/26954538

#SPJ1

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