Answer:
(a): 46 degrees
(b): 63 degrees
(c): 27 degrees
(d): 60 degrees
Step-by-step explanation:
If period of

is one-half the period of

and
<span>

has a period of 2π, then

and

.
</span>
To find the period of sine function

we use the rule

.
<span /><span />
f is sine function where f (0)=0, then c=0; with period

, then

, because

.
To find a we consider the condition

, from where

.
If the amplitude of

is twice the amplitude of

, then

has a product factor twice smaller than

and while period of

<span> </span> is 2π and g(0)=0, we can write

.
Let's divide the shaded region into two areas:
area 1: x = 0 ---> x = 2
ares 2: x = 2 ---> x = 4
In area 1, we need to find the area under g(x) = x and in area 2, we need to find the area between g(x) = x and f(x) = (x - 2)^2. Now let's set up the integrals needed to find the areas.
Area 1:

Area 2:





Therefore, the area of the shaded portion of the graph is
A = A1 + A2 = 5.34
Answer:
169.56
Step-by-step explanation:
54*3.14 = 169.56
this is the answer on the test
Answer:
1.7689 (rounded to 4 decimal places)
Step-by-step explanation:
Let the number we are seeking be "x", thus we can write the equation as:

Since we have raised "x" to the "10th power", to get "x" back again, we need to take 10th root. Same goes for right side, we take 10th root of 300. We will get our answer. The process shown below:
![x^{10}=300\\\sqrt[10]{x^{10}} =\sqrt[10]{300} \\x=\sqrt[10]{300} \\x=1.7689](https://tex.z-dn.net/?f=x%5E%7B10%7D%3D300%5C%5C%5Csqrt%5B10%5D%7Bx%5E%7B10%7D%7D%20%3D%5Csqrt%5B10%5D%7B300%7D%20%5C%5Cx%3D%5Csqrt%5B10%5D%7B300%7D%20%5C%5Cx%3D1.7689)
hence, 1.7689 raised to 10th power will give us 300