This is a 3-4-5 triangle.
Remember,
SOH-CAH-TOA
S(ine) = O(pposite)/H(ypotenuse)
C(osine) = A(djacent)/H(ypotenuse)
T(angent) = O(pposite)/A(djacent)
In Cosine, the adjacent = 3, and the Hypotenuse = 5
Hypotenuse is the longest side.
3,4,5 is the triangle that fits inside, therefore, the Opposite is 4, and Adjacent is 3
T(angent) = O(pposite)/A(djacent)
Plug in the numbers
T = 4/3
tan = 4/3 is your answer
hope this helps
Answer: none
Step-by-step explanation:
(A)
(16÷32/10) ×2 + 0.2×(90)
Using bodmas principle ; solve bracket
(16×10/32)×2 + (2/10×90)
10+18 =28
(B)
{(16÷32/10) × (2+2/10)} ×90
Open brackets
{(16×10/32) × (22/10)} ×90
(5×11/5) ×90
11×90 = 990
(C)
16÷{(32/10×2) + (2/10×8)} +82
Open brackets, solve division first, dolled by addition
16÷(32/5 + 8/5) +82
16÷(40/5) +82
16÷8 +82
2+82= 84
(D)
[16÷(32/10 ×2) + 0.2× (90)]
16÷ (32/5) + 2/10 ×90
Solve division
16×5/32 + 18
5/2 + 18
L.c.m of denominator (2&1) =2
(5+36) / 2 = 41/2
=20.5
Answer: Choice A) Add 3.8 to both sides of the equation
Explanation:
If we knew the value of w, then we would replace it and apply PEMDAS.
However, we don't know the value of w, so we undo each step of PEMDAS going backwards.
We start with the "S" of PEMDAS, and undo the subtraction. To undo subtraction, you apply addition. To undo that "minus 3.8" we add 3.8 to both sides.
Answer:
The answer is 6
Step-by-step explanation:
Same as plane surfaces on a regular cuboid
