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3 years ago

Me need more help plz also add explanation real explanation

Mathematics

2 answers:

Reika [66]3 years ago

8 0

So we want to find the surface area of this pyramid. Finding the surface area of a 3D shape is essentially finding the area of all its sides and adding them together.

What we're going to have to do here is find the area of the base of the pyramid, which is basically a square, and find the area of all four sides of the pyramid, which are triangles.

Formula for area of a square: length × width

Formula for area of a triangle: (base × height) ÷ 2

First we will find the area of the base:

l × w = A

(Write out formula)

8 × 8 =

(Input values)

A = 64 m^2 (square meters or meters squared)

(We have our answer, dont forgot to write your units.)

So we've figured out the area of the base. Now onto the triangles:

(b×h) ÷ 2 = A

(Write out formula)

(8 × 6) ÷ 2 =

(Input values)

48 ÷ 2 =

(Use PEMDAS to solve this correctly. We did whatever was in the parenthesis first, which was multiplication.)

A = 24m^2 (square meters or meters squared)

(We have our answer.)

Now we've found the area of the base of the pyramid and one of the sides of the pyramid, but we're almost done! There's four sides to the pyramid, they're all the same. So since we found the area for one, we can multiply that by 4 to find the area of all sides added together:

24 × 4

(area of a side × total sides)

= 96m^2

All that's left is adding the 96 to the area of the base to get our total surface area of the pyramid:

96 + 64 = 160

The surface area of the pyramid is 160m^2(square meters or meters squared)

(Hope this helps :) )

Lisa [10]3 years ago

8 0

Answer:okkkkkkkk

Step-by-step explanation:360

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a)19.28≈x (rounded to the nearest hundredth)

b)7.66≈x (rounded to the nearest hundredth)

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I'm going to first assume that all of these are right triangles.

If you look at the hint, it says SOHCAHTOA

This stands for the trigonometric functions

SOH tells us that sine(θ)=opposite/hypotenuse

CAH tells us that cosine(θ)=adjacent/hypotenuse

TOA tells us that tangent(θ)=opposite/adjacent

(ps. θ, or theta, stands for any angle measure)

Now, let's look at problem a.

We have the value of the hypotenuse and the value of an angle. We want to find the side opposite from the angle. So, in this case, we want to use the sine function.

sin(40)=x/30

30*sin(40)=x

19.28≈x (rounded to the nearest hundredth)

We can solve the other problems in a similar fashion.

Problem b

We have the hypotenuse, an angle measure, and we want to find the adjacent side. This is telling us to use the cosine function.

cos(40)=x/10

10*cos(40)=x

7.66≈x (rounded to the nearest hundredth)

Problem c

We have an angle measure, an adjacent value, and we want to find the opposite length, so we're going to use the tangent function.

tan(40)=x/10

10*tan(40)=x

x≈8.39(rounded to the nearest hundredth)

Problem d

We need to first solve for x, the angle measure.

We have an opposite length, an adjacent length, so we can use the tangent function to solve for the missing angle

Let's set up an equation

tan(x)=3/4

Using $tan^{-1}$ or tangent inverse to cancel out the tangent, we get

x= $tan^{-1}$ (3/4)

x≈36.87 (rounded to the nearest hundredth)

Now, let's solve for y, the hypotenuse!

Both the sine and cosine functions work, so I'm just going to pick one.

sine(36.87)=3/y

y*sin(36.87)=3

y=3/sin(36.87)

y=5

Side note: Although I rounded the angle measure, when calculating the y value, on my calculator, I did not round the value. Don't round until the end on your calculator, but it's okay to round on your paper.

PS. I used degrees for the angle measures, not radians, so if you want to retry these problems, make sure your calculator is calculating degrees, not radians, or else you might get some funky answers.

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