Answer:
Part 1) m∠1=45°
Part 2) m∠3=45°
Part 3) m∠2=135°
Part 4) m∠4=135°
Step-by-step explanation:
we know that
m∠1=m∠3 -----> by vertical angles Equation A
m∠2=m∠4 -----> by vertical angles Equation B
m∠1+m∠2=180° ----> by supplementary angles (linear pair) Equation C
3(m∠1+m∠3) = m∠2+m∠4 ----> Equation D
Substitute equation A and equation B in equation D
3(m∠1+m∠1) = m∠2+m∠2
6(m∠1) = 2m∠2
3(m∠1) =m∠2 -----> equation E
Substitute equation E in equation C and solve for m∠1
m∠1+3(m∠1)=180°
4(m∠1)=180°
m∠1=45°
<em>Find the measure of m∠3</em>
Remember that
m∠1=m∠3 (equation A)
therefore
m∠3=45°
<em>Find the measure of m∠2</em>
Remember that
m∠2=3(m∠1) (equation E)
substitute the value of m∠1
m∠2=3(45°)=135°
<em>Find the measure of m∠4</em>
Remember that
m∠2=m∠4 (equation B)
therefore
m∠4=135°
Answer:
The slope of the line passing through the points (-4, 8) and (2,8) is 0
Step-by-step explanation:
The slope is 0 because you would subtract the y ones first 8-8, subtract the x ones next 2-(-4). You would get 0/6. When you would divide 0/6, you would get 0
The fare of $(20 - 2.5) = $17.5 will maximize the total fare.
<h3>What is Differentiation?</h3>
Differentiation means the rate of change of one quantity with respect to another. The speed is calculated as the rate of change of distance with respect to time.
Here, The operator for a round-trip fare of $20, carries an average of 500 people per day.
It is estimated that 20 fewer people will take the trip, for each $1 increase in fare.
for $x increase in fare, 20x less people will take the trip and at that time the total fare F is given by
f(x) =(20 + x)(500 - 20x)
f (x) = 10000 + 100x - 20x²
For f(x) to be maximum, the condition is dy/dx = 0
100 - 40x = 0
⇒ x = 2.5
Thus, the fare of $(20 - 2.5) = $17.5 will maximize the total fare.
Learn more about Differentiation from:
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Answer:
9/10 I do this a lot
step-by-step explanation:
1/2 changes to 5/10 and 2/5 becomes 4/10 then you add and there is your answer
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π and cos A = cos B · cos C
scratchwork:
A + B + C = π
A = π - (B + C)
cos A = cos [π - (B + C)] Apply cos
= - cos (B + C) Simplify
= -(cos B · cos C - sin B · sin C) Sum Identity
= sin B · sin C - cos B · cos C Simplify
cos B · cos C = sin B · sin C - cos B · cos C Substitution
2cos B · cos C = sin B · sin C Addition
Division
2 = tan B · tan C
<u>Proof LHS → RHS</u>
Given: A + B + C = π
Subtraction: A = π - (B + C)
Apply tan: tan A = tan(π - (B + C))
Simplify: = - tan (B + C)
Substitution: = -(tan B + tan C)/(1 - 2)
Simplify: = -(tan B + tan C)/-1
= tan B + tan C
LHS = RHS: tan B + tan C = tan B + tan C