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liraira [26]
3 years ago
12

You are packing a boxed gift in a larger shipping box. The dimensions of the shipping box are all 1.5 times the dimensions of th

e boxed gift. The volume of the shipping box is 1080 cubic inches. What volume of the shipping box is not taken up by the gift box?
Mathematics
1 answer:
omeli [17]3 years ago
6 0
If the dimensions of the shipping box are all 1.5 times those of the gift, then the shipping box is 1.5*1.5*1.5, or 3.375 times larger than the gift. If the shipping box is 1080 cu. in., then the gift is 1080/3.375, or 320 cu.in.. Therefore; the amount of unused space is 1080-320, or 760
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Step-by-step explanation:

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Write the following expression using symbolic language.
Ad libitum [116K]

Answer:

<h3>              1)   7x - 5 </h3><h3>              2)   9y - 18 </h3><h3>              3)   0.5n + 4n </h3><h3>              4)  2(w³+23)</h3><h3> Step-by-step explanation:</h3>

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The product of seven and a number x:   7·x = 7x

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<h3>7x - 5 </h3>

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nine times a number y:  9·y = 9y

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<h3>9y - 18 </h3>

3)

half a number n:   0.5n

four times the number:   4·n = 4n

<u>Half a number n increased by four times the number:</u>

<h3>0.5n + 4n </h3>

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a number w cubed:  w³

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5 0
3 years ago
(x^4+15x^3-77x^2+13x-36)+(x-4)
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7 0
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Read 2 more answers
​Find all roots: x^3 + 7x^2 + 12x = 0 <br> Show all work and check your answer.
Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

x^{3}+7 x^{2}+12 x=0

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

x^{3}+7 x^{2}+12 x=0

x\left(x^{2}+7 x+12\right)=0   ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

x^{2}+7 x+12=0

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}

<em><u>Therefore, the two roots are:</u></em>

\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}

And,

\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0
3 years ago
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