PLEASE HELP GIVING BRAINLIEST

Find 2004-04-02-04-00_files/i0340000.jpg. Round the answer to the nearest tenth.

lesya692 [45]

<span>The problem is to calculate the angles of the triangle. However, it is not clear which angle you have to calculate, so we are going to calculate all of them
</span>
we know that
Applying the law of cosines
c²=a²+b²-2*a*b*cos C------> cos C=[a²+b²-c²]/[2*a*b]
a=12.5
b=15
c=11
so
cos C=[a²+b²-c²]/[2*a*b]---> cos C=[12.5²+15²-11²]/[2*12.5*15]
cos C=0.694------------> C=arc cos (0.694)-----> C=46.05°-----> C=46.1°

applying the law of sines calculate angle B
15 sin B=11/sin 46.1-----> 15*sin 46.1=11*sin B----> sin B=15*sin 46.1/11
sin B=15*sin 46.1/11-----> sin B=0.9826----> B=arc sin (0.9826)
B=79.3°

calculate angle A
A+B+C=180------> A=180-B-C-----> A=180-79.3-46.1----> A=54.6°

the angles of the triangle are
A=54.6°
B=79.3°
C=46.1°

7 0

3 years ago

A jumbo crayon is composed of a cylinder with a conical tip. The cylinder is 12 cm tall with a radius of 1.5 cm, and the cone ha

s2008m [1.1K]

Answer:

Part 1) The lateral area of the cone is $LA=2\pi\ cm^{2}$

Part 2) The lateral surface area of the cylinder is $LA=36\pi\ cm^{2}$

Part 3) The surface area of the crayon is $SA=41.50\pi\ cm^{2}$

Step-by-step explanation:

Part 1) Find the lateral area of the cone

The lateral area of the cone is equal to

$LA=\pi rl$

we have

$r=1\ cm$

$l=2\ cm$

substitute

$LA=\pi (1)(2)$

$LA=2\pi\ cm^{2}$

Part 2) Find the lateral surface area of the cylinder

The lateral area of the cylinder is equal to

$LA=2\pi rh$

we have

$r=1.5\ cm$

$h=12\ cm$

substitute

$LA=2\pi (1.5)(12)$

$LA=36\pi\ cm^{2}$

Part 3) Find the surface area of the crayon

The surface area of the crayon is equal to the lateral area of the cone, plus the lateral area of the cylinder, plus the top area of the cylinder plus the bottom base of the crayon

<em>Find the area of the bottom base of the crayon</em>

$A=\pi[r2^{2}-r1^{2}]$

where

r2 is the radius of the cylinder

r1 is the radius of the cone

substitute

$A=\pi[1.5^{2}-1^{2}]$

$A=1.25\pi\ cm^{2}$

<em>Find the area of the top base of the cylinder</em>

$A=\pi(1.5)^{2}=2.25\pi\ cm^{2}$

<em>Find the surface area</em>

$SA=2\pi+36\pi+2.25\pi+1.25\pi=41.50\pi\ cm^{2}$

8 0

3 years ago

Read 2 more answers

Many people are surprised to find that math is found in the Bible in many places. God created all things and sustains all things

uranmaximum [27]

Answer:

Solid 1:

T.A. = 2(wl + hl + hw)

T.A. = 2(2 * 8 + 4 * 8 + 4 * 2)

T.A. = 2(16 + 32 + 8)

T.A. = 32 + 64 + 16

T.A. = 96 + 16

T.A. = 112m^2

V = w * h * l

V = 2 * 4 * 8

V = 8 * 8

V = 64m^3

M = 0.003 * 64

M = $0.19

Solid 2:

T.A. = l sqr(l^2 + 4 * h^2) + l^2

T.A. = 6 sqr(6^2 + 4 * 5^2) + 6^2

T.A. = 106m^2

V = 1/3 * b * h

V = 1/3 * 36 * 5

V = 12 * 5

V = 60m^3

M = 0.003 * 60

M = $0.18

Solid 3:

T.A. = (2 * pi * r^2) + (2 * pi * r * h)

T.A. = (2 * pi * 2.5^2) + (2 * pi * 2.5 * 3.5)

T.A. = (2 * pi * 6.25) + (2 * pi * 2.5 * 3.5)

T.A. = 39.27 + 54.98

T.A. = 94.25m^2

V = pi * r^2 * h

V = pi * 2.5^2 * 3.5

V = pi * 6.25 * 3.5

V = pi * 21.88

V = 68.72m^3

M = 0.003 * 68.72

M = $0.21

Solid 4:

T.A. = pi * r^2 + pi * r * sqr(r^2 + h^2)

T.A. = pi * 4^2 + pi * 4 * sqr(4^2 + 6^2)

T.A. = pi * 16 + pi * 4 * sqr(16 + 36)

T.A. = 50.26 + 12.56 * sqr(52)

T.A. = 62.82 * sqr(52)

T.A. = 140.9m^2

V = 1/3 * pi * r^2 * h

V = 1/3 * pi * 4^2 * 6

V = 1/3 * pi * 16 * 6

V = 1/3 * pi * 96

V= pi * 32

V = 100.5m^3

M = 0.003 * 100.5

M = $0.30

Solid 5:

T.A. = 4 * pi * r^2

T.A. = 4 * pi * 2.6^2

T.A. = 4 * pi * 6.76

T.A. = pi * 27.04

T.A. = 84.95m^2

V = 4/3 * pi * r^3

V = 4/3 * pi * 2.6^3

V = 4/3 * pi * 17.576

V = pi * 23.434

V = 73.62m^3

M = 0.003 * 73.62

M = $0.22

Write a paragraph that answers the following questions:

● Do all of the solids hold between 60 and 70 cubic inches?

● Which solid is the most cost efficient (the packaging with the smallest materials cost that holds between 60 and 70 cubic inches)?

● Would the Ark be most cost efficient solid for packaging? Why or why not?

● Which solid would you recommend using? Why?

● Why do you think that God was very specific in the measurements that He gave Noah for the Ark? Be sure to use some of your findings from this project to support your answer.

Of the five shapes, only the rectangular prism, the square pyramid, and the cylinder held between 60 and 70 cubic inches. The square pyramid was the most cost efficient, holding a price of $0.18. A rectangular prism would not be the most efficient shape cost wise, however it would be sufficient to hold the materials. I think I would recommend the rectangular prism for this project, because it is the second most cost efficient and holds a greater volume for use. I think God gave Moses the exact plans needed to perfectly fit all his animals. I think he wanted to prove to Moses his power and knowledge by providing the exact amount needed. In this project I took the measurements and I used them to discover which was best suited to serve the purpose needed, but God already knew and used it as a lesson in faith for Moses.

Step-by-step explanation:

7 0

2 years ago

A square pyramid is sliced in half, parallel to the base.

DIA [1.3K]

Answer:

I think the answer is a because it not getting any smaller

7 0

2 years ago

Read 2 more answers

Factor 2x^2-4x-30 completely. A. 2(x-5)(x+3) B. (2x-5)(x+3) C. (2x+3)(x-5) D. 2(x-3)(x+5)

OLEGan [10]