Answer:
m∠x = 10°
Step-by-step explanation:
Since we are dealing with a <u>right triangle</u> and its two sides and one angle, we can use <u>trigonometry ratios</u>. Remember them all with the acronym SohCahToa.
"o" is for opposite side, "a" is for adjacent side, "h" is for hypotenuse side.
The ratios are:
sinθ = opposite/hypotenuse
cosθ = adjacent/hypotenuse
tanθ = opposite/adjacent
θ means the "angle of reference", or the angle you know or want to find. This determines which side is adjacent (touching) or opposite (not touching). The hypotenuse (longest side) does not change.
In this triangle, θ = x. The sides we know are hypotenuse and opposite. Therefore, we will use the sinθ ratio.
sinθ = opposite/hypotenuse
sinx = CB/AB Substitute the labels in the diagram
sinx = 4/23 Substitute known values (side lengths)
x = sin⁻¹(4/23) Isolate 'x'. Use calculator to solve.
x = 10.015....° Exact answer
x ≈ 10° Round to the nearest degree
Therefore the measure of angle x (m∠x) is 10°.
Answer:
16, see below :)
Step-by-step explanation:
Hello!
We can use the pythagorean theorem because one of the angles is 90 degrees. According to the theorem, a^2 + b^2 = c^2
a is 12, b is unknown, and c is 20
Now we use the theorem
12^2 + b^2 = 20^2
144 + b^2 = 400
b^2 = 256
b = sqrt(256)
b = 16
Assuming you mean f(t) = g(t) × h(t), notice that
f(t) = g(t) × h(t) = cos(t) sin(t) = 1/2 sin(2t)
Then the difference quotient of f is

Recall the angle sum identity for sine:
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
Then we can write the difference quotient as

or

(As a bonus, notice that as h approaches 0, we have (cos(2h) - 1)/(2h) → 0 and sin(2h)/(2h) → 1, so we recover the derivative of f(t) as cos(2t).)
Answer:
I think so its neither
Step-by-step explanation:
I dont know
Answer:

Step-by-step explanation:
<u><em>The correct question is</em></u>
1x10 to the fifth power over 4x10 to the negative fourth power given on scientific notation
we know that
To divide two numbers in scientific notation, divide their coefficients and subtract their exponents.
we have
