Brainliest for anyone if they get this CORRECT. Check ALL that apply.

A regular triangular pyramid has a base and lateral faces that are congruent equilateral triangles. It has a lateral surface are

TEA [102]

We know that

[lateral surface area of a regular triangular pyramid]=3*[area of one equilateral triangle]
so
[area of one equilateral triangle]=lateral surface/3-----> 81/3-----> 27 ft²

[surface area of the regular triangular pyramid]=lateral area+area of the base

area of the base is equals to the area of the lateral sides because are equilateral triangles

therefore
area of the base=27 ft²

[surface area of the regular triangular pyramid]=81+27----> 108 ft²

the answer is
the surface area of the regular triangular pyramid is 108 ft²

4 0

2 years ago

Solve for the unknown by using the additive inverse. Type the FULL answer in the box, without using any spaces (ex., X=5).

zavuch27 [327]

Answer:

X=16

Step-by-step explanation:

10x-8=9x+8

First you have to simplify.

10x-9x=1x

So now the equation looks like this

x-8=8

Now you have to get x by itself

-8+8=0 and we have to do the same thing on both sides of the equal sign 8+8=16

The equation now looks like this x=16

And that's your answer

3 0

3 years ago

Find the volumes of the solids generated by revolving the triangle with vertices (2, 2), (2, 6), and (5, 6) about a) the x-a

Vesna [10]

About the $x$ -axis (washer method):

$\displaystyle\pi\int_2^5\left(6^2-\left(\frac43x-\frac23\right)^2\right)\,\mathrm dx=\frac{16\pi}9\int_2^5(20+x-x^2)\,\mathrm dx=\boxed{56\pi}$

About the $y$ -axis (shell method):

$\displaystyle2\pi\int_2^5x\left(6-\left(\frac43x-\frac23\right)\right)\,\mathrm dx=\frac{8\pi}3\int_2^5x(5-x)\,\mathrm dx=\boxed{36\pi}$

About $x=7$ (shell method):

$\displaystyle2\pi\int_2^5(7-x)\left(6-\left(\frac43x-\frac23\right)\right)\,\mathrm dx=\frac{8\pi}3\int_2^5(35-12x+x^2)\,\mathrm dx=\boxed{48\pi}$

About $y=2$ (washer method):

$\displaystyle\pi\int_2^5\left((6-2)^2-\left(\frac43x-\frac23-2\right)^2\right)\,\mathrm dx=\frac{16\pi}9\int_2^5(5+4x-x^2)\,\mathrm dx=\boxed{32\pi}$

7 0

3 years ago

How do I do this and what’s the answer

lakkis [162]

Answer:

The area of the four curved surfaces is the area of one circle with a radius of 4 cm

Area = PI * radius^2

Area = PI * 4 ^ 2

Area = PI * 16 = 50.26548245744 sq cm

Area of entire square = 8 * 8 = 64 sq cm

Area of yellow region = 64 - 50.265 sq cm =

= 13.735 = 13.7 sq cm (rounded to nearest tenth)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The value of y varies directly with x and inversely with

Pani-rosa [81]

Answer:

x = 45 when y = 20 and z = 6

Step-by-step explanation:

The general formula here is y = kx/z²

If x = 24 and y = 6, z = 8, so:

y = kx/z² becomes 6 = k(24)/8², and:

6 1

k = ---------------- = -------------- = 16

24/64 4/64

Then y = 16x/z², and so, if y = 20 and z = 6 (check this!), then:

20 = 16x/36, or 16x = 720, or x = 45

x = 45 when y = 20 and z = 6

3 0

2 years ago