Given:
The figures of triangles and their mid segments.
To find:
The values of n.
Solution:
Mid-segment theorem: According to this theorem, mid segment of the triangle is a line segment that bisect the two sides of the triangle and parallel to third side, The measure of mid-segment is half of the parallel side.
9.
It is given that:
Length of mid-segment = 54
Length of parallel side = 3n
By using mid-segment theorem for the given triangle, we get
Divide both side by 3.
Hence, the value of n is equal to 36.
10.
It is given that:
Length of mid-segment = 4n+5
Length of parallel side = 74
By using mid-segment theorem for the given triangle, we get
Divide both side by 4.
Hence, the value of n is equal to 8.
<span>Ok. You would set up the problem like this: 32/x=64/100 and then cross multiply; 3200=64x; divide both sides by 64; x=50. Mary has 50 customers.</span>
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ
Use the Sum/Difference Identities:
sin(α + β) = sinα · cosβ + cosα · sinβ
cos(α - β) = cosα · cosβ + sinα · sinβ
Use the Unit circle to evaluate: sin45 = cos45 = √2/2
Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ
Use the Pythagorean Identity: cos²Ф + sin²Ф = 1
<u />
<u>Proof LHS → RHS</u>
LHS: 2sin(45 + 2A) · cos(45 - 2A)
Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)
Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]
Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]
Distribute: cos²2A + 2cos2A·sin2A + sin²2A
Pythagorean Identity: 1 + 2cos2A·sin2A
Double Angle: 1 + sin4A
LHS = RHS: 1 + sin4A = 1 + sin4A 
Answer:
There is no diagram for our observation. Please refine your question next time.
Step-by-step explanation:
Answer:
x= -4
Step-by-step explanation:
Using BODMAS, bracket first, multiply all in the bracket with -2, getting the answer, collect like terms and the answer is gotten
