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Alex787 [66]
2 years ago
8

A semi-ellipse-shaped gate has a maximum height of 20 feet and a width of 15 feet, both measured along the center. Can a truck w

ith a height of 12 feet and a width of 16 feet with a load and 10 feet without a load pass through the gate?
Mathematics
1 answer:
Free_Kalibri [48]2 years ago
6 0
<h2>Answer with explanation:</h2>

We are given a semi-ellipse gate whose dimensions are as follows:

  Height of 20 feet and a width of 15 feet.

Now, if a truck is loaded then:

Height of truck is: 12 feet and a width of truck is: 16 feet

The truck won't pass through the gate since the width of truck is more than that of the gate.

When the truck is not loaded then:

 Height of truck is: 12 feet and a width of truck is: 10 feet

The truck would easily pass through  the gate since, the dimensions of truck are less than that of the gate.

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Coffee with sugar and milk added to it can contain up to calories. To consume a maximum of 600 mg if caffeine per day, how many
Marina CMI [18]
We know there is 330 mg of caffeine per 16-ounce serving. So if we want 8-ounce servings:
330 mg : 2 = 165 mg
Also we have to consume a maximum of 600 mg.
165 mg * 3 = 495 mg < 600 mg
165 mg * 4 = 660 mg > 600 mg
Answer: Someone could drink 3 full 8-ounce servings in a day and still stay below the limit.
8 0
2 years ago
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Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv
Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

   = (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = \dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}

Let dD/dT = 0 to find the critical value we will get

\dfrac{70000000\left(291T^2-24298T+91800\right)}  = 0

Using formula of quadratic, we get the roots:

T =  79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8)    a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0
2 years ago
Rob is making a scale model of the Solar System on the school field. He wants the distance from the Sun to Jupiter to be 8 metre
ANEK [815]

Answer:

1 : 9.75 * 10⁷

Step-by-step explanation:

To find n, we have to divide the real distance by the scale distance. This is 7.8 * 10⁸ / 8 = 0.975 * 10⁸ = 9.75 * 10⁷ which means that the ratio is 1 : 9.75 * 10⁷.

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2 years ago
Two forces, F1 and F2, are represented by vectors with initial points that are at the origin. The first force has a magnitude of
Strike441 [17]

Answer:

Step-by-step explanation:

The two force F1 and F2  are represented by vectors with initial points that are at the origin.

the terminal point of the vector is point P(1, 1, 0)

Therefore, the direction of the vector force is

v_1=(1-0)\hat i+(1-0)\hat j +(0-0)\hat k\\\\=1\hat i+1\hat j+0\hat k\\\\=\hat i + \hat j

The unit vector in the direction of force will be

\frac{v_1}{|v_1|} =\frac{\hat i+ \hat j}{\sqrt{1^2+1^2+0^2} } \\\\=\frac{1}{\sqrt{2} (\hat i +\hat j)}

The magnitude of the force is 40lb, so the force will be

F_1=40\times \frac{1}{\sqrt{2} } (\hat i+\hat j)\\\\=20\sqrt{2} (\hat i+\hat j)

The terminal point of its vector is point Q(0, 1, 1)

Therefore, the direction of the vector force is

v_2=(0-0)\hat i+(1-0)\hat j +(1-0)\hat k\\\\=0\hat i+1\hat j+1\hat k\\\\=\hat j + \hat k

\frac{v_2}{|v_2|} =\frac{\hat j+ \hat k}{\sqrt{0^2+1^2+1^2} } \\\\=\frac{1}{\sqrt{2} (\hat j +\hat k)}

The magnitude of the force is 60lb, so the force will be

F_2=60\times \frac{1}{\sqrt{2} } (\hat j+\hat k)\\\\=30\sqrt{2} (\hat j+\hat k)

The resultant of the two forces is

F=F_1+F_2\\\\=[20\sqrt{2} (\hat i+\hat j)]+[30\sqrt{2} (\hat j +\hat k)]\\\\=20\sqrt{2} \hat i+20\sqrt{2} \hat j +30\sqrt{2} \hat j+30\sqrt{2} \hat k\\\\=20\sqrt{2} \hat i+50\sqrt{2} \hat j+30\sqrt{2} \hat k

The magnitude force will be

|F|=\sqrt{(20\sqrt{2} )^2+(50\sqrt{2} )^2+(30\sqrt{2} )^2} \\\\=\sqrt{800+5000+1800} \\\\=\sqrt{3100} \\\\=55.68

to (1 decimal place)=55.7lb

b) The direction angle of force F

The angle formed by F and x axis

\alpha=\cos^{-1}(\frac{20\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.5080)\\\\=59.469

The angle formed by F and y axis

\alpha=\cos^{-1}(\frac{50\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(1.270)\\\\=

The angle formed by F and z axis

\alpha=\cos^{-1}(\frac{30\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.7620)\\\\=40.359

8 0
1 year ago
Now evaluate f(x) = 2x4 - 4x3-11x2+3x-6 for x=-2 f (-2) =
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f(x)=2x^4-4x^3-11x^2+3x-6\\\\f(-2)=2\cdot(-2)^4-4\cdot(-2)^3-11\cdot(-2)^2+3\cdot(-2)-6\\\\f(-2)=2\cdot16-4\cdot(-8)-11\cdot4+3\cdot(-2)-6\\\\f(-2)=32+32-44-6-6=8
4 0
2 years ago
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