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3 years ago

Factor the polynomial 16b^4+30b^3-12b^2

Mathematics

1 answer:

Lelu [443]3 years ago

4 0

Answer:

2b²(8b²+15b-6)

Step-by-step explanation:

16 $b^{4}$ +30b³-12b²

2b²(8b²+15b-6)

This is not further factorable. If you want to find the roots, you'd have to use the quadratic formula for the polynomial in the parentheses. Therefore that is our final answer.

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Help, > < or =? |-6| ? 4

Damm [24]

Answer:

Step-by-step explanation:

Because |-6| = 6

and 6 is greater than 4

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3 years ago

Which side lengths form a right triangle 4, 8, 12

Lostsunrise [7]

(4^2+8^2)^.5 = 8.944
It's not 12.
So 4,8,12 can't be the side lengths of a right triangle.

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3 years ago

A survey about the art program at a school found that 250 students wanted an after-school art program, 150 students wanted art c

Harman [31]

250 out of 500 students is going to give you 0.5. After that you will need to multiply that number by 360. your equation will look something like this: 0.5 x 360 = 180
your welcome! :)

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3 years ago

The choir director set up chairs for the concert with an equal number of rows and columns.

Dima020 [189]

'y' is NOT the number of chairs in each row and column. It's the TOTAL number of
chairs, so it's the product of (chairs in each row) times (chairs in each column).
The problem says that those numbers are equal. So we would expect the TOTAL
to be a perfect square, and the number in each row and each column to be the
square root of the total. Since the total is 81, the number in each row and column
could be either +9 or -9 . Both answers are mathematically correct, but -9
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5 0

4 years ago

Read 2 more answers

The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s

lys-0071 [83]

Answer:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

$IQR = Q_3 -Q_1 = 250-210 =40$

For this case we need to find the lower and upper limit with the following formulas:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

4 0

4 years ago