Answer:
<em>The uniform rate of depreciation is $56,333 per year, or 11% per year.</em>
Step-by-step explanation:
<u>Rate of Depreciation</u>
If some object has a certain value P and after some time t its value decreases to Q, the rate of depreciation is a measure of how much its value changed per unit time.
It can be calculated as:
It can also be expressed as a percentage, dividing the previous value by P.
The school bus reduces its value from $512,000 to $343,000 in 3 years, thus:
R = 56,333 $/year
Expressed as a percentage:
R = 56,333 / 512,000 = 0.11
R = 11%/year
The uniform rate of depreciation is $56,333 per year, or 11% per year.
Looking at the data it looks like a line. So the answer will be in y+mx+b format.
Knowing it is a line eliminates answers b and c.
looking at options a is says it starts at about .1 when x is zero. We can see this looks pretty close as the left-most dot is below 1 already.
The slope on a is 1/2--> increasing 1 vertically for every 2 horizontally. The pattern mostly follows that rule.
Therefore a (y=1/2x+.1) would be the correct answer
Answer:
Step-by-step explanation:
<u>Trigonometric Ratios</u>
The ratios of the sides of a right triangle are called trigonometric ratios.
The longest side of the triangle is called the hypotenuse and the other two sides are the legs.
The sine ratio relates the opposite side of an angle of the triangle with the hypotenuse:
Here, the opposite leg to angle B is the side CA=2, and the hypotenuse is the side AB=3, thus:
Calculating the inverse sine function:
Answer:
1. Transversal y intersects lines m and n; <1 ~= <2 (Given)
2. <1 ~= <3 (Vertical Angles Theorem)
3. <2 ~= <3 (Transitive Property of Congruence)
4. m || n (Converse of Alternate Interior Angles Theorem)
Answer:
Step-by-step explanation:
Given
See attachment
From the attachment, we have:
First, we need to calculate length LM,
Using Pythagoras theorem:
Collect Like Terms
Solving (a): 
Substitute values for MN and LN
Solving (b): 
Substitute values for LM and MN
Solving (c): 
Substitute values for LN and LM
Solving (d): 
Substitute values for LM and LN
