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Semenov [28]
3 years ago
14

Jenna wants to paint a wooden jewelry box that is 8 inches by 6 inches by 3 inches with clear paint. The bottle of paint says it

will cover a total area of 250 square inches. Does she have enough to cover the entire box​
Mathematics
1 answer:
Tanzania [10]3 years ago
7 0
Yes. 6x8 equals 48. 48x3 equals 144.
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Approximate the real number solution(s) to the polynomial function f(x) = x^3 + 4x^2 + x − 6.
irakobra [83]

Answer:

d. x ∈ {-3, -2, 1}

Step-by-step explanation:

The word "approximate" in the problem statement suggests that the roots may not be integers. Consequently, the solution method of choice is graphing.

A graph shows the solutions to f(x)=0 to be x=-3, x=-2, and x=1, matching selection D.

_____

<em>Comment on the problem statement</em>

Any value of x can be put into the formula and it will be a "solution." For example, f(0) = -6 is a "solution" to the polynomial function. The problem statement is not specific as to the solution(s) sought. We have had to guess that we want values of x such that f(x)=0.

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4 years ago
Convert the decimal expansion 0.291666... to a fraction.
NNADVOKAT [17]
Pa arti tu fraction
7 0
4 years ago
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Factorise 3x^3 - 12xy^2​
igomit [66]

Answer:

we can conclude that:

3x^3\:-\:12xy^2=3x\left(x+2y\right)\left(x-2y\right)

Step-by-step explanation:

Given the expression

3x^3-12xy^2

Let us factorize the expression

3x^3-12xy^2

Apply the exponent rule: a^{b+c}=a^ba^c

3x^3\:-\:12xy^2=3xx^2-12xy^2

Rewrite 12 as 4 · 3

                     =3xx^2-4\cdot \:3xy^2

Factor out the common term 3x

                     =3x\left(x^2-4y^2\right)

\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)

                      =3x\left(x+2y\right)\left(x-2y\right)

Therefore, we can conclude that:

3x^3\:-\:12xy^2=3x\left(x+2y\right)\left(x-2y\right)

8 0
3 years ago
Call an integer $n$ oddly powerful if there exist positive integers $a$ and $b$, where $b&gt;1$, $b$ is odd, and $a^b = n$. How
azamat

Answer:

There are 16 oddly powerful integers less than 2010

Step-by-step explanation:

∵ b is an odd integer

∵ b > 1

∴ The first value of b is 3

∵ a is an integer

- We can use a = 1, 2, 3, ..........

∵ a^{b}=n

∵ n < 2010

- Let a = 1, 2, ............... 12 because 12³ is greatest integer  < 2010

∵ 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125, 6³ = 216, 7³ = 343,

   8³ = 512, 9³ = 729, 10³ = 1000, 11³ = 1331, 12³ = 1728

∴ There are 12 oddly powerful integers with b = 3

Now the second value of b is 5

1^{5}=1 but we took 1 before so we will start with 2

∵ 2^{5}=32, 3^{5}=243, 4^{5}=1024

- 4^{5} is the greatest integer < 2010

∴ There are 3 oddly powerful integers with b = 5

Now the third value of b is 7

∵ 2^{7}=128

- 2^{7} is the greatest integer < 2010

∴ There is 1 oddly powerful integers with b = 7

Now the fourth value of b is 9

∵ 2^{9}=512

- 2^{9} is the greatest integer < 2010

- But we used 512 before

∴ There is no oddly powerful integers with b = 9

- 9 is the greatest value of b which makes a^{b}

∵ 12 + 3 + 1 = 16

∴ There are 16 oddly powerful integers less than 2010

5 0
3 years ago
A certain rollercoaster can accomodate 1,600 riders per hour. A digital sign states there are 345 people in line. If you get int
sdas [7]
1 hour I'm pretty sure
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3 years ago
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