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Serggg [28]
3 years ago
5

The length of a rectangular prism is three times as long as its width, and its height is two feet longer than its width. If the

volume of the prism is 135 cubic feet, what is its width?
Mathematics
1 answer:
aivan3 [116]3 years ago
7 0
A rectangular prism is a polyhedron with six rectangular faces. To fully define a rectangular prism, one must know its length, width and height. The volume of a rectangular prism is expressed as the product of its dimensions - length, width and height. To calculate for its width, we use the volume formula.

V = l x w x h
From the given, we have:
l = 3w
h = 2 + w
V = 135 ft^3

Substituting the given expressions,
135 ft^3 = 3w (w) (2+w)
135 ft^3 = 6w^2 + 3w^3

Solving for w, we obtain the following values:
w1= 3
w2 = 2.5 + 2.96i
w3 = 2.5 - 22.96i

Since width should not have a negative value, the value of w should be 3 ft.
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Find the real solution(s) of the equation. Round your answer to two decimal places when appropriate. 5x^6=30
Elina [12.6K]

Answer:

The real solutions are

x=\sqrt[6]{6}\approx 1.35\\\\\:x=-\sqrt[6]{6}\approx -1.35

Step-by-step explanation:

The solution, or root, of an equation is any value or set of values that can be substituted into the equation to make it a true statement.

To find the real solutions of the equation 5x^6=30:

\mathrm{Divide\:both\:sides\:by\:}5\\\\\frac{5x^6}{5}=\frac{30}{5}\\\\\mathrm{Simplify}\\\\x^6=6\\\\\mathrm{For\:}x^n=f\left(a\right)\mathrm{,\:n\:is\:even,\:the\:solutions\:are\:}x=\sqrt[n]{f\left(a\right)},\:-\sqrt[n]{f\left(a\right)}\\\\x=\sqrt[6]{6}\approx 1.35\\\\\:x=-\sqrt[6]{6}\approx -1.35

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3x² - 7 x 2 ? É isso ? 
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4 years ago
Please help me with this
Ne4ueva [31]

Answer:

Option c

Step-by-step explanation:

A system of equations are given to us. And we need to solve them . The given system is

\begin{cases} y = 2x - 3\dots 1 \\ y = x^2 - 3\dots 2 \end{cases}

We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .

\implies y = x^2 - 3 \\\\\implies 2x - 3 = x^2 - 3 \\\\\implies x^2 - 2x = 0 \\\\\implies x(x-2) = 0 \\\\\implies\red{ x = 0 , 2 }

We got two values of x as 0 & 2 . Alternatively substituting these values we have ,

\implies y = 2 x - 3 \\\\\implies y = 2(0)-3 \qquad or \qquad y = 2(2)-3 \\\\\implies y = 0-3 \qquad or \qquad 4 - 3 \\\\\implies \red{ y = -3 , 1 }

Thefore the required answer is ,

\red{Option\:c} \begin{cases} (0,-3) \\ (2,1) \end{cases}

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