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3 years ago

Please help me on this I’m super stuck

Mathematics

1 answer:

erastova [34]3 years ago

6 0

Answer:

Step-by-step explanation:

To solve this you need to know that cosine∅ = adjacent/hypotenuse. We already know that hypotenuse = 5 and that ∅ = 70*. By substituting these in you will get cosine 70* = adjacent/5. Evaluate cosine 70* on the calculator and get 0.633... Multiply both sides by 5 and get adjacent is about equal to 3.1666. Because BC is the adjacent side of the triangle, this is the length of BC.

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Find the midpoint of (1,8) and (9,6)

Anon25 [30]

Answer:

(5,7)

Step-by-step explanation:

(1,8)

(9,6)

x+x= 9+1=10

y+y= 8+6=14

ans. divide by 2

(5,7)

4 0

2 years ago

How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

shepuryov [24]

$LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \tan { x } } \\ \\ =\frac { 1 }{ \tan { x } } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } -\frac { \sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \right) \\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x-\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } }{ \cos ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\left( 1-\cos ^{ 2 }{ x } \right) } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\cdot \sin ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } \sin ^{ 4 }{ x } }{ \sin { x\cos ^{ 2 }{ x } } } \\ \\ =\frac { \sin ^{ 3 }{ x } }{ \cos { x } }$

$\\ \\ =\sin ^{ 2 }{ x } \cdot \frac { \sin { x } }{ \cos { x } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \frac { \cos { x } }{ \sin { x } } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \cot { x } } \\ \\ =\frac { \sin ^{ 2 }{ x } }{ \cot { x } } \\ \\ =RHS$

8 0

2 years ago

Find the measure of the angle formed between the base of the cone and a line segment that represents the slant height

wel

Since there are no figures given, I will give an example.

You are given a silo that is shaped as a closed cylinder with a conical end. The diameter of the silo is 4 ft, the length of the cylindrical part is 6 ft, and the entire length of the silo is 10.5 ft. Suppose that you are asked to find the total volume of grains that can be stored in the silo.

Given:

Cylinder part

D = 4 ft

H = 6 ft

Cone part

H = 10.5 – 6 = 4.5ft

D = 4ft

Required:

Volume of silo

Solution:
V of cylinder = πr²H

V of cylinder = π(4/2)²(6)

V of cylinder = 75.4 ft³

V of cone = πr²H/3

V of cone = π(4/2)²(4.5/3)

V of cone = 18.85 ft³

Total volume = 94.25 ft³

6 0

2 years ago

Write the equation of a line that is parallel to y=0.6x+3 and that passes through the point

a_sh-v [17]

A line parallel to y = 0.6x + 3 will be of the form:

$y = 0.6x + k$

where k is a constant.

It passes through (-3,-5)

Thus,

$y = 0.6x + k \\ \\ -5 = 0.6 \times (-3) + k \\ \\ -5 = -1.8 + k \\ \\k = -5 + 1.8 \\ \\ k = -3.2$

$\textbf{The equation is \: \: y = 0.6x - 3.2}$

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2 years ago

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AlexFokin [52]

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6 0

2 years ago