Answer:
steps below
Step-by-step explanation:
3.2.1 AD = DB* sin 2 = DB * sin θ .. DE // AB ∠2= θ ... (1)
By laws of sines: DB / sin ∠5 = x / sin ∠4
∠4 = θ-α ∠5 = 180°-<u>∠1</u>-∠4 = 180°-<u>∠3</u>-∠4 = 180°-(90°-θ)-(θ-α)) = 90°+α
DB = (x*sin ∠5)/sin (θ-α)
= (x* sin (90°+α)) / sin (θ-α)
AD = DB*sinθ
= (x* sin (90°+α))*sinθ / sin (θ-α)
= x* (sin90°cosα+cos90°sinα)*sinθ / sin (θ-α) .... sin90°=1, cos90°=0
= x* cosα* sinθ / sin (θ-α)
3.2.2 Please apply Laws of sines to calculate the length
The value of the composite function is (f/g)(5)= 26
<h3>How to evaluate the
composite function?</h3>
The functions are given as:
f(x) = x^2+1 and g(x) = x-4
To calculate the composite function, we use the following formula
(f/g)(x)= f(x)/g(x)
So, we have:
(f/g)(5)= f(5)/g(5)
Calculate f(5) and g(5)
f(5) = 5^2+1 and g(5) = 5-4
This gives
f(5) = 26 and g(5) = 1
Substitute f(5) = 26 and g(5) = 1 in (f/g)(5)= f(5)/g(5)
(f/g)(5)= 26/1
Evaluate
(f/g)(5)= 26
Hence, the value of the composite function is (f/g)(5)= 26
Read more about composite function at
brainly.com/question/10687170
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6 bills in the deposit were $10 bills.
<u>Step-by-step explanation:</u>
The total number of bills he gave the teller = 24 bills
x + y = 24 ------------(1)
Let,
x be the number of $5 bills.
y be the number of $10 bills.
The total bank deposit made using $5 and $10 bills are $150.
<u>He deposits $5 bills and $10 bills. So, the equation can be framed as :
</u>
5x + 10y = 150 --------(2)
The question is asked for how many $10 bills were in the deposit,
Here 'y' is the number of $10 bills.
<u>To solve for y value :
</u>
The eq(1) must be multiplied by 5 and subtract eq(2) from eq(1),
5x + 5y = 120
-(<u>5x + 10y = 150</u>)
<u>-5y = -30 </u>
⇒ y = 30/5
⇒ y = 6
Therefore, there were 6 bills in the deposit that are $10 bills.
The coordinate point is 8
A measure of the variation around the estimate of the mean.
The standard deviation is a statistic that lets you know how tightly all the subjects are clustered around the mean in a set of data. A single standard deviation away from the mean in either of the directions accounts for somewhere around 68 % of data. If two standard deviations are away from the mean, it accounts for around 95% of data. If three standard deviations are away it accounts for 99% of the data.
1) The total range of variation in the dataset is called RANGE.
3) A measure of the variation around the estimate of the mean is SEM (STANDARD ERROR OF THE MEAN).
4) The most common value observed (highest frequency) is MODE.
