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Crazy boy [7]
3 years ago
14

Find the approximate side length of a square game board with an area of 164 in2

Mathematics
1 answer:
notsponge [240]3 years ago
4 0
Squares have the same length and width.
Area of a rectangle/square is length X width
A=LW
164 = LW
Now we know L and W are the same because its a square.
13X13 is 169
12X12 is 144
It appears the lengths are just less than 13.
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Please help, i need this done asap!
Bond [772]

Answer:

}\frac{\sqrt{2} }{2} or 0.707

Step-by-step explanation:

cos(315) is equivalent to cos (\frac{7\pi }{4})

You can look at the unit circle to find the cos value for this. Keep in mind that cos is the x value of the corresponding coordinate.

7 0
3 years ago
Help please and thank u
elena55 [62]

Answer:

It’s correct

Step-by-step explanation:

You‘ve got the first angle correct! You‘ve got the second angle correct. The third angle is also correct! angle.

Now the third angle has a way of Tricken ya! So heres a way to solve:

5 x 20= 100

100-40= 60

GOOD JOB! have a good night!

3 0
3 years ago
Which one????????????
VARVARA [1.3K]
Which what ? Put the question and will help you
6 0
3 years ago
A truncated cone has horizontal bases with radii 18 and 2. A sphere is tangent to the top, bottom, and lateral surface of the tr
MariettaO [177]
<h3>Answer:  6 units</h3>

===================================================

Explanation:

A truncated cone is where we start with a regular 3D cone and chop off the top. The portion up top is a smaller cone, which we'll ignore. The bottom part is the truncated cone portion.

In the real world, a lampshade is one example of a truncated cone.

----------------

Let x be the radius of the sphere.

We'll be focusing on a vertical cross section of the truncated cone. Refer to figure 1 in the diagram below.

We have the following points

  • A = center of the sphere
  • B = center of the circular base of the truncated cone
  • C = point 18 units to the right of point B
  • E = point directly above point A, and its the center of the circular top of the truncated cone
  • D = point 2 units to the right of point E
  • F = the location where the circle touches the slanted curved side of the truncated cone
  • G = point directly below point D, and located on segment BC

We'll connect a few of those points forming the dashed lines in figure 1.

To start off, draw a segment from D to G. This forms rectangle BGDE with sides of EB = 2x and BG = 2. The side BC is 18 units, so that must mean GC = BC - BG = 18 - 2 = 16.

Since EB = 2x, this means DG is also 2x.

------------------

Now focus on triangles ABC and AFC. They are congruent right triangles. We can prove this using the HL (hypotenuse leg) theorem. Recall that the radius of a circle is perpendicular to the tangent line, which is why angle AFC is 90 degrees. Angle ABC is a similar story.

Because they are congruent right triangles, this indicates side BC is the same length as side FC. Therefore, FC = 18

Through similar logic, triangles ADE and ADF are congruent as well which leads to ED = FD = 2.

Combine sides FD and FC to get the length of DC

DC = FD+FC = 2+18 = 20

This is the hypotenuse of the right triangle GCD

------------------

After all that, we have the right triangle GCD with the legs of 2x and 16. The hypotenuse is 20. Refer to figure 2 shown below.

As you can probably guess, we'll use the pythagorean theorem to find the value of x.

a^2 + b^2 = c^2

(DG)^2 + (GC)^2 = (CD)^2

(2x)^2 + (16)^2 = (20)^2

4x^2 + 256 = 400

4x^2 = 400-256

4x^2 = 144

x^2 = 144/4

x^2 = 36

x = sqrt(36)

x = 6 is the radius of the sphere.

7 0
2 years ago
An oil tank has to be drained for maintenance. The tank is shaped like a cylinder that is 4 ft long with a diameter of 2.2 ft. S
gavmur [86]

Answer:

8\:\mathrm{minutes}

Step-by-step explanation:

The volume of a cylinder is given as:

r^2\cdot\pi\cdot h

Since the diameter is 2.2, the radius is \frac{2.2}{2}=1.1.

Therefore, the volume of this cylinder is:

1.1^2\cdot 3.14\cdot 4=15.1976\:\mathrm{ft^3}, using \pi=3.14 as requested in the problem.

Since the tank is emptied at a constant rate of 1.5\:\mathrm{ft^3\:per\:minute}, it will take

\frac{15.1976}{1.5}=7.5988=\fbox{$8\:\mathrm{minutes}$} to empty the tank.

8 0
3 years ago
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