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kirill115 [55]
4 years ago
6

Bricks are delivered to a work site and stacked in rows and columns, forming a rectangular prism. The length of the prism is 3 f

eet greater than its width, and its height is 1 foot less than its width. Find the dimensions of the prism formed by the bricks, given that its volume is 1170 cubic feet.
(I set it up, and I've tried solving it, and this is all I have so far. If you could help me out by completing it AND explaining so I can better understand it, that would be great!)
Here's what I have so far:
W= x, L= x+3, H= x-1
Volume= 1170 ft^3, so L*W*H= 1170 ft^3

(x+3)*x*(x-1) (after this is where I get stuck)
10 points to whoever answers correctly! Thank you for your time!!!

Mathematics
1 answer:
topjm [15]4 years ago
6 0
Correct so far. Finish up.
.. (x +3)(x)(x -1) = 1170
.. x^3 +2x^2 -3x -1170 = 0 . . . . . . one sign change ⇒ one positive real solution

This is a cubic equation with one positive real solution. There are several possible approaches to solving this. My favorite is to let a graphing calculator show me the solution.

You can also factor 1170 = 13*10*9, so x=10.

The width of the prism is 10 feet; the length is 13 feet; the height is 9 feet.

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an electrician has 4.1 meters of wire. How much stripes 7/10m long can he cut? How much wire will he have left over?
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Answer:

5 such strips of \frac{7}{10}\ m can be cut and \frac{6}{10}\ m  would be left over.

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(HELP URGENT RESPOND QUICKLY) What is the correct classification of 32?
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A triangle has three sides 35cm 54 cm and 61 cm find its area Also find the smallest of its altitude ​
I am Lyosha [343]

Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

  • ( a ) = 35 cm

  • ( b ) = 54 cm

  • ( c ) = 61 cm

We can find the area of the triangle with its three sides using Heron's Formula

  • <u>Heron's </u><u>Formula</u>

Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

\sf{   \pmb { \longrightarrow \:  \sqrt{s(s - a)(s - b)(s - c)} }}

where ( s ) :

\sf \longrightarrow s = \dfrac{a + b + c}{2}

Therefore, for the given triangle first we will calculate ( s )

\begin {aligned}\quad & \quad \longmapsto  \sf  s =  \dfrac{a + b + c}{2}  \\  & \quad \longmapsto  \sf s =  \dfrac{35 + 54 + 61}{2}  \\ & \quad \longmapsto  \sf s =  \dfrac{150}{2}  \\ & \quad \longmapsto  \sf s  = 75cm \end{aligned}

Now, Area of triangle will be:

\begin{aligned}&:\implies \sf\quad \sf \:  A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \:  A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)}   \\&:\implies \sf\quad \sf \:  A = \sqrt{75 \times 40 \times 21 \times 14}  \\ &:\implies \sf\quad \sf \:  A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5}  \\ &:\implies \sf\quad \sf \:  A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5}  \\ &:\implies \sf\quad \sf \:  A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf   A =939.15 {cm}^{2} }} \end{aligned}

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

\begin{aligned}&:\implies \sf\quad \sf \:  Area =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2}  \times b \times h =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2} \times 61 \times h  = 939.15 \\&:\implies \sf\quad \sf \:  h =939.15 \times  \dfrac{2}{61}   \\&:\implies \sf\quad \sf \:  h = \dfrac{1818.3}{61}  \\ &:\implies \sf\quad  \boxed{ \pmb{\sf \:  h =30.79 \: (approx)}} \end{aligned}

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