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frutty [35]
3 years ago
5

A paving company was hired to make a 4 mile section of the highway. They need 700 tons of concrete to complete the job. How many

pounds of concrete will they need to repave the highway?
Mathematics
1 answer:
Arlecino [84]3 years ago
5 0

Answer:

they will need 1,400,000

Step-by-step explanation:

1 ton is 2,000 pounds. so 700x 2,000=1,400,000

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Laura bought a notebook and pencil for $4.50 the notebook cost twice the amount of the pencil
skad [1K]

Answer: N=3

Step-by-step explanation:

Let's represent notebooks with "N" and pencils with "P".

Then,

N + P = 4.50

Since N =2P, we can substitute that directly into the equation:

(2P) + P = 4.50

3P = 4.50 (divide both sides by 3)

P = 1.50

Now that we have P, we can in turn substitute that into the equation.

N + (1.50) = 4.50

N = 3.00

7 0
3 years ago
Find the maximum and minimum values of the curve y=2x³-3x²-12x+10​
Ray Of Light [21]

\underline{  \orange{\huge \boxed{ \frak{Answer : }}}}

Let ,

\sf \large \color{purple} y = 2 {x}^{3}   - 3 {x}^{2}  - 12x + 10  \: --( \: 1 \: )

\: \: \:

Now , Diff wrt ' x ' , we get :

\sf \:  \frac{dy}{dx}  =  \frac{d}{dx} (2 {x}^{3}  - 3 {x}^{2}  - 12x + 10) \\  \sf \: \sf \:  \frac{dy}{dx}  =  \frac{d}{dx} \: 2(3 {x}^{2} ) -  \frac{d}{dx} 3 {x}^{2}  -  \frac{d}{dx} 12x +  \frac{d}{dx} 10 \\  \sf \: \frac{dy}{dx}  =2(3 {x}^{2} ) - 3(2x) - 12(1) + 0 \\  \sf \: \frac{dy}{dx}  =6 {x}^{2}  - 6x - 12 + 0 \\  \: \sf \red{\frac{dy}{dx}  = 6 {x}^{2} 6x - 12 -- (2)}

\: \: \:

For maxima or minima \frac{dy}{dx} = 0

\: \: \:

\sf \: 6 {x}^{2}  - 6x - 12 = 0

\: \: \:

Divided by 6 on both side , we get.

\: \: \:

\sf \:  {x}^{2}  - x - 2 = 0 \\  \sf \:  {x}^{2}  - 2x + x - 2 = 0 \\  \sf \: x(x - 2) + 1(x - 2) = 0 \\  \sf \: (x - 2)(x + 1) = 0 \\  \sf \: x - 2 = 0 \:  \:  \bold or \:  \: x + 1 = 0 \\ \sf \fbox{x = 2 \: }  \: \bold or \:  \fbox{ x =  - 1}

\: \: \:

Again Diff wrt ‘ x ’ , we get.

\sf \: \frac{d}{dx}  =(\frac{dy}{dx}  ) = 6\frac{d}{dx}  - 6\frac{d}{dx}x - \frac{d}{dx}12 \\  \sf \:  \frac{ {d}^{2}y }{ {dx}^{2} }  = 6(2x) - 6(1) - 0 \\ \sf \: \sf \bold{  \frac{ {d}^{2}y }{ {dx}^{2} }  =12x - 6}

\: \: \:

At x = 2

\: \: \:

\sf \:  \frac{ {d}^{2}y }{ {dx}^{2} }  =12(2) - 6 \\  \:  \:  \:  \sf \:  = 24 - 6 \\   \:  \:  \:  \: \sf \red{  = 18  > 0}

At x = -1

\: \: \:

\sf \:  \frac{ {d}^{2}y }{ {dx}^{2} }  =12( - 1) - 6  \\   \:  \:  \: \sf \:   = - 12 - 6 \\  \:  \:  \:  \: \sf \red{  =  - 18 < 0 }

\: \: \:

x = 2 gives minima value of function.

\: \: \:

x = -1 gives maxima value of function.

\: \: \:

Now, put x = 2 in eqⁿ ( 1 )

\: \: \:

\sf \: y \: minima  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = 2( {2})^{3}  - 3 ({2})^{2}  - 12(2) + 10 \\  \:  \:  \:   \:  \:  \: \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \:  \:  \:   \sf \:   \:  \:   \:  = 2(8) - 3(4) - 24 + 10 \\ \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \sf \:   \:  \:  \:  \:  = 16  - 12 - 24 + 10 \\\sf \:   \:  \:  \:  \:  \:  \:  \:  \:  \: =  - 20 + 10 \\\sf \color{red}{\boxed{  =  - 10}}

\: \: \:

<u>The </u><u>Point </u><u>of </u><u>minima </u><u>is </u><u>(</u><u> </u><u>2</u><u> </u><u>,</u><u> </u><u>-</u><u>1</u><u>0</u><u> </u><u>)</u><u>.</u>

\: \: \:

Now , put x = -1 in eqⁿ ( 1 )

\sf \: y \: maxima  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   = 2( { - 1})^{3}  - 3 ({ - 1})^{2}  - 12( - 1) + 10  \\\sf  \color{red}{\boxed{  =  17}}

\: \: \:

<u>The point of maxima value is ( -1 , 17 )</u><u>.</u>

\: \: \:

<u>\:  \:</u>

Hope Helps! :)

5 0
2 years ago
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vladimir2022 [97]
So first you convert
2 and 5/8 and
1 and 1/2
2 and 5/8=16/8+5/8=21/8
1 and 1/2=2/2+1/2=3/2
so 21/8-3/2
convert so that the denominator (bottom number) is the same
21/8-([3/2][4/4])
21/8-12/8
9/8=1 and 1/8
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