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lapo4ka [179]
2 years ago
14

You are given a rectangular sheet of cardboard that measures 11 in. by 8.5 in. (see the diagram below). A small square of the sa

me size is cut from each corner, and each side folded up along the cuts to from a box with no lid.
1. Anya thinks the cut should be 1.5 inches to create the greatest volume, while Terrence thinks it should be 3 inches.

Explain how both students can determine the formula for the volume of the box.

Determine which student's suggestion would create the larger volume.

Explain how there can be two different volumes when each student starts with the same size cardboard.

2. Why is the value of x limited to 0 in. < x < 4.25 in.?

Mathematics
1 answer:
GaryK [48]2 years ago
7 0

Answer:

Terrence's

Step-by-step explanation:

The length of the square that will be cut out is the height of the box.

1. a

Anya's method: 8.5 -1.5 =7, 11- 1.5 =9.5, the height is 1.5, so the volume is height x length x width which is 1.5 x 9.5 x 7 =99.75 squared inches.

Terrence's method: 8.5-3 = 5.5, 11-3 = 8. Vol= 5.5 x 8 x 3 =132 squared inches. 99.75 < 132 squared inches, Terrence's idea would create larger volume.

1. b

The box's size depends on the length/width/height of the cardboard being cut, which is why different measurements / cutting methods for the same size cardboard can result in different box sizes.

2. The square would be cut from all four corners, therefore the sum of the 2 squares on the cardboard cannot exceed the short side of the cardboard. The shorter side of the cardboard is 8.5 inches, divided by 2 = 4.25 inches, hence the squares cannot  be larger than 4.25 inches. Keep in mind that if you cut exactly 4.25 inches you will have a strip of 2.5 inches width that cannot be turned into a box.

If you want to cut 5 inches squares out, depending on how you draw it, it would either overlap or go outside of the paper because 5+5 is ten, surely on the 11 inches side that would still be perfectly fine but for the 8.5 inches side, there isn't any room for the 10 inches.

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What is the length of ac?
leonid [27]

Answer:

Step-by-step explanation:

Given triangles are similar, so we setup a proportion  

(140-x)/81 = x/9

solving gives x = 14

AC = 140-x = 140-14 = 126

7 0
2 years ago
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Sales tax in one state is 4%. What is the amount of tax on a $56.70 purchase?
NeTakaya

1) Convert 4% to decimal = .04

2) Multiply the purchase by that decimal

.04(56.70) = $2.27

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5 0
2 years ago
8.5b−3.4∙(13a−3.2b)+a
Nesterboy [21]

Answer:

8.5b - 3.4(13a - 3.2b) + a = 19.4b - 43.2a

Step-by-step explanation:

It is a simple mathematical problem with multiple like terms. We can solve it by applying basic mathematical rules of multiplication and addition/subtration.

8.5b - 3.4(13a - 3.2b) + a

= 8.5b - 3.4*13a -3.4*(-3.2b) + a

= 8.5b - 3.4*13a + 3.4*3.2b + a

= 8.5b - 44.2a + 10.88 b + a

Now, only like terms can be added to each other

= (8.5b + 10.9b) + (a - 44.2a)

= 19.4b + (-43.2a)

= 19.4b - 43.2a    

8 0
2 years ago
A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon
posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x

(concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)

Let y<em>(t)</em> be the amount of salt (in pounds) in the tank at time <em>t</em> (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)}  \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)}  \:\frac{pound}{s}

V(t) is the volume of brine in the tank at time <em>t. </em>To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

V(t)=100 + 2t

We can then write the initial value problem:

\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

\frac{dy}{dx}+P(x)y =Q(x)

where <em>P</em> and <em>Q</em> are continuous functions on a given interval.

In our case, we have that

\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, \mu \left( x \right), using \mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}

\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}

Step 2: Multiply everything in the differential equation by \mu \left( x \right) and verify that the left side becomes the product rule \left( {\mu \left( t \right)y\left( t \right)} \right)' and write it as such.

\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}

Step 3: Integrate both sides.

\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C

Step 4: Find the value of the constant and solve for the solution y(t).

50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000

y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}

Now, the tank is full of brine when:

V(t) = 400\\100+2t=400\\t=150

The amount of salt in the tank when it is full of brine is

y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75

6 0
2 years ago
The infield of a baseball field is a square
AnnZ [28]

Answer:

  106.1 ft/s

Step-by-step explanation:

You know the diagonal of a square is √2 times the length of one side, so the distance from 3rd to 1st is 90√2 feet ≈ 127.2792 ft.

The speed is the ratio of distance to time:

  speed = distance/time = 127.2972 ft/(1.2 s) ≈ 106.1 ft/s.

_____

In case you have never figured or seen the computation of the diagonal of a square (the hypotenuse of an isosceles right triangle), consider the square with side lengths 1. The diagonal will cut the square into halves that are isosceles right triangles with leg lengths 1. Then the Pythagorean theorem can be used to find the diagonal length d:

  d² = 1² + 1²

  d² = 2

  d = √2

Since this is the diagonal for a side length of 1, any other side length will serve as a scale factor for this value. A square with a side length of 90 ft will have a diagonal measuring 90√2 ft.

7 0
2 years ago
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