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3 years ago

What is the surface area of the square pyramid?

Mathematics

2 answers:

zmey [24]3 years ago

8 0

The answer is d 288 cm2

nika2105 [10]3 years ago

3 0

The area is simply , the area of 4 triangles + area of base

area of triangle = 0.5 * 6 * 8 = 24 cm^2

area of triangle = 4 * 24 + (8)^2= 160 cm^2

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A certain city has a population of about 100,000. There are 3,101 people with red hair in the city. Which of these is closest to

Delicious77 [7]

Answer:

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Step-by-step explanation:

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2 years ago

Solve the equation for x: x/3 - 6 = x/5 - 4

lorasvet [3.4K]

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2 years ago

4(1-x)+2x=-3(x+1) -) 7/5 -) -7/5 -) -3/4 -) -7

natima [27]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

What is the coefficient in the expression 7x+14

Lady_Fox [76]

Answer:

Step-by-step explanation:

The coefficient is the number that comes before the variable

7x+14

In this case, the variable is x.

The number that comes before x is 7, so 7 is the coefficient.

Hope this helps! :)

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2 years ago

Read 2 more answers

A plane flying horizontally at an altitude of 3 miles and a speed of 500 mi/h passes directly over a radar station. Find the rat

konstantin123 [22]

Answer:

The rate at which the distance from the plane to the station is increasing is 331 miles per hour.

Step-by-step explanation:

We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:

a: is one side of the triangle = altitude of the plane = 3 miles

b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles

h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles

First, we need to find b:

$a^{2} + b^{2} = h^{2}$ (1)

$b = \sqrt{h^{2} - a^{2}} = \sqrt{(4 mi)^{2} - (3 mi)^{2}} = \sqrt{7} miles$

Now, to find the rate we need to find the derivative of equation (1) with respect to time:

$\frac{d}{dt}(a^{2}) + \frac{d}{dt}(b^{2}) = \frac{d}{dt}(h^{2})$

$2a\frac{da}{dt} + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

Since "da/dt" is constant (the altitude of the plane does not change with time), we have:

$0 + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

And knowing that the plane is moving at a speed of 500 mi/h (db/dt):

$\sqrt{7} mi*500 mi/h = 4 mi*\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{\sqrt{7} mi*500 mi/h}{4 mi} = 331 mi/h$

Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.

I hope it helps you!

4 0

2 years ago