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Mamont248 [21]
3 years ago
9

A general contractor is constructing a building that requires a concrete foundation that is to be 20 feet by 22 feet and 24 inch

es thick. If the local home supply store sells concrete for $ 110 per cubic yard, what will be the cost of the concrete for the foundation?
Express your answer rounded up to the next whole dollar.
Mathematics
2 answers:
Aneli [31]3 years ago
6 0

Answer: $2,613,600

Step-by-step explanation:

Concept:

Here, we need to know the idea of unit conversion and a rectangular prism.

When there are multiple units occurring in the same question, then we should convert all the units into one by different conversion factors.

A rectangular prism is a three-dimensional figure that has 6 sides that are rectangles.

1 foot = 12 inches

1 yard = 3 feet

V (rectangular prism) = w · l · h

Solve:

24 inches = 24 / 12 = 2 feet

$110 per cubic yard = 110 × 3³ = $2970 per cubic feet

V = w · l · h

V = (20) (22) (2)

V = 880 feet³

880 × 2970 = $2,613,600

Hope this helps!! :)

Please let me know if you have any questions

ad-work [718]3 years ago
3 0

Answer:

$38427

Step-by-step explanation:

Surface Area Formula: A=2(wl+hl+hw) where l=length w=width h=height

A=2((20(22))+(22(2))+(20(2)) = 1048ft^2

3 ft = 1yds -> 1048 ft = 349 1/3 yds

349 1/3(110)= 38426.66666

$38427

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Which statement is correct regarding the measurements of the parallelogram?
Mekhanik [1.2K]

Answer:

The base is 8 and the height is 3, so the area is 8 (3) = 24 square units.

Step-by-step explanation:

The area of the parallelogram is given by the following expression:

A = \|\vec u\times \vec v\|

The vectors are, respectively:

\vec u = (10-2, 1 - 1,0-0)

\vec u = (8,0,0)

The base of the parallelogram is 8 units.

\vec v = (8-2, 4-1,0-0)

\vec v = (6,3,0)

The height of the parallelogram is 3 units.

The cross product of both vectors is:

\vec u \times \vec v = (0,0,24)

The area of the parallelogram is given by the norm of the resulting vector:

\|\vec u \times \vec v\| = 24

8 0
3 years ago
Read 2 more answers
The table shows the function representing the height in the area of the base of a rectangular prism for different values of X th
stepan [7]

Answer:

A

<em>can i have brainliest lol</em>

Step-by-step explanation:

when x = 3, the volume is 36.

4 times 9 = 36, so you multiply f(x) by g(x) to get 36

6 0
3 years ago
Factorise completely:<br> (a) 8m - 50 n2
Sever21 [200]

Answer:

2(4m-25n2)

Step-by-step explanation:

To factorise,we will have to know the value that can divide both variables given

But based on this question,we have different variables that is m and n

So we don't have to deal with that but with the numbers which means that we will only deal with the numbers ignoring the letters

8m-50n2

So the only number that can divide the two is 2

2(4m-25n2)

This cannot be solved further so the final answer is 2(4m-25n2)

I hope this will help you

7 0
4 years ago
A man bought 42 stamps, some 13¢ and some 18¢. How many of each kind did he buy if the cost was $6.66?
yawa3891 [41]
Solution:

Here we have 2 system:
1. x + y = 42
2. 13x + 18y = 666

From equation 1.:
3. y = 42 - x

Substitute equation 3 into equation 2.:
13x + 18(42 - x) = 666
-5x + 756 = 666
5x = 90
x = 18

Since we got the value of x, then we will find the value of y. Substitute x into Equation 3:
y = 42 - 18 = 24
4 0
4 years ago
Solve the following biquadratic equation:<br><br> <img src="https://tex.z-dn.net/?f=%28%287%21%29%2Ax%5E4%29%2B%28%285%21%29%2Ax
jekas [21]
\large\begin{array}{l} \textsf{Solve the equation}\\\\ \mathsf{7!\cdot x^4+5!\cdot x^2-3=0}\\\\ \mathsf{7!\cdot (x^2)^2+5!\cdot x^2-3=0}\\\\\\ \textsf{Substitute}\\\\ \mathsf{x^2=t\quad(t\ge 0)}\\\\\\ \textsf{so the equation becomes}\\\\ \mathsf{7!\cdot t^2+5!\cdot t-3=0}\quad\Rightarrow\quad\left\{\! \begin{array}{l} \mathsf{a=7!}\\\mathsf{b=5!}\\\mathsf{c=-3} \end{array} \right. \end{array}


\large\begin{array}{l} \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(5!)^2-4\cdot 7!\cdot (-3)}\\\\ \mathsf{\Delta=(5!)^2+12\cdot 7!}\\\\ \mathsf{\Delta=5!\cdot 5!+12\cdot 7!}\\\\ \mathsf{\Delta=5!\cdot 5!+12\cdot 7\cdot 6\cdot 5!}\\\\ \mathsf{\Delta=5!\cdot (5!+12\cdot 7\cdot 6)}\\\\ \mathsf{\Delta=5!\cdot (120+504)}\\\\ \mathsf{\Delta=5!\cdot 624} \end{array}

\large\begin{array}{l} \mathsf{\Delta=120\cdot 2^4\cdot 3\cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3\cdot 5)\cdot 2^4\cdot 3\cdot 13}\\\\ \mathsf{\Delta=2^{3+4}\cdot 3\cdot 5 \cdot 3\cdot 13}\\\\ \mathsf{\Delta=2^7\cdot 3^2\cdot 5 \cdot 13} \end{array}

\large\begin{array}{l} \mathsf{\Delta=2^{6+1}\cdot 3^2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=2^{6}\cdot 2\cdot 3^2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=2^{3\,\cdot\,2}\cdot 3^2\cdot 2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3)^2\cdot 2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3)^2\cdot 130} \end{array}


\large\begin{array}{l} \mathsf{t=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{t=\dfrac{-(5!)\pm \sqrt{(2^3\cdot 3)^2\cdot 130}}{2\cdot 7!}}\\\\ \mathsf{t=\dfrac{-120\pm 2^3\cdot 3\sqrt{130}}{2\cdot 7!}}\\\\ \mathsf{t=\dfrac{-2^3\cdot 3\cdot 5\pm 2^3\cdot 3\sqrt{130}}{2\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3!}}\\\\ \mathsf{t=\dfrac{2^3\cdot 3\cdot \big(\!\!-5\pm \sqrt{130}\big)}{2\cdot 4\cdot 7\cdot (2\cdot 3)\cdot 5\cdot 3!}}\\\\ \mathsf{t=\dfrac{2^3\cdot 3\cdot \big(\!\!-5\pm \sqrt{130}\big)}{2\cdot 2^2\cdot 3\cdot 7\cdot 2\cdot 5\cdot 3!}} \end{array}

\large\begin{array}{l} \mathsf{t=\dfrac{(2^3\cdot 3)\cdot \big(\!\!-5\pm \sqrt{130}\big)}{(2^3\cdot 3)\cdot 7\cdot 2\cdot 5\cdot 3!}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{7\cdot 2\cdot 5\cdot 3\cdot 2\cdot 1}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{2^2\cdot 7\cdot 5\cdot 3\cdot 1}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{2^2\cdot 105}} \end{array}

\large\begin{array}{l} \begin{array}{rcl} \mathsf{t=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}&~\textsf{ or }~& \end{array} \mathsf{t=\dfrac{-5-\sqrt{130}}{2^2\cdot 105}}\quad\textsf{(useless, since }\mathsf{t\ge 0}\textsf{)}\\\\ \mathsf{t=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}} \end{array}


\large\begin{array}{l} \textsf{Substituite back for }\mathsf{x^2=t:}\\\\ \mathsf{x^2=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}\\\\ \mathsf{x=\pm \sqrt{\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}}\\\\ \mathsf{x=\pm \dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}}\\\\ \boxed{\begin{array}{rcl} \mathsf{x=-\,\dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}}&~\textsf{ or }~&\mathsf{x=\dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}} \end{array}} \end{array}


\large\begin{array}{l} \textsf{Solution: }\mathsf{S=\left\{-\,\frac{1}{2}\sqrt{\frac{-5+\sqrt{130}}{105}},\,\frac{1}{2}\sqrt{\frac{-5+\sqrt{130}}{105}}\right\}.} \end{array}


If you're having problems understanding the answer, try to see it through your browser: brainly.com/question/2094277


\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best wishes! :-)} \end{array}


Tags: <em>solve biquadratic equation factorial Bhaskara solution</em>

4 0
3 years ago
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