1. the complement is when two angles add up to 99 degrees and the supplement is when the two angles add up to 180 degrees.
2. Let x be the measure of the unknown angle .
3. Let 90 -x be it's complement and 180- x be the supplement of the angle .
4. Given 90-x=3x+9 the measure of the complement of an angle is 3x+9. Solve using the following steps.
99-x=3x+9
Add x-9 to both sides to isolate x-terms on right side and numbers on left
90-x+x-9=3x+9+x-9
90-x+x-9=3x+9+x-9
91=4x
divide by 4 to both sides yields
91/4=x
5. Given 3x+99=180-x. The measure of the supplement of an angle is 3x+99. So
3x+99=180-x
Add x+99 to both sides of equations to isolate x on right side and number on left
3x+99+x-99=180-x+x-99
4x=91
X=91/4
6. The angle is x=91/4
The number of books sold is 473.
<u>Step-by-step explanation:</u>
- The original cost of each book = $0.64
- The selling price of each book = $0.75
The difference between the original price and selling price of the book gives the profit per book.
The profit of one book = Selling price - Original price
Let,
- The total number of books be 'x'.
- The number of books sold be 'y'.
- The unsold books is 100.
- The total profit is -12 because it was gone to a loss of $12.
Therefore, the equation is formed as
total Profit = 0.75y - 0.64x
⇒ 0.75y - 0.64x = -12 --------(1)
Total books = sold books + unsold books
x = y + 100
⇒ x-y = 100 -------(2)
Substitute x= 100+y in the eq(1),
0.75y - 0.64(100+y) = -12
0.75y - 64 -0.64y = -12
0.11y = -12 +64
y = 52 / 0.11
y = 472.7
y ≅ 473
The number of book sold is 473 books.
The total number of books is (100+473) = 573 books.
Find the rate of change, (y2-y1)/(x2-x1), from the data given...
(2.16-1.26)/(12-7)=0.18
(1.26-0.72)/(7-4)=0.18
Since the rate is constant, this is a linear equation of the form y=mx+b. Furthermore, since 0 pencils cost 0, b=0, so the cost of the pencils is simply the number of pencils times 18 cents...
c(p)=0.18p (cost with respect to pencils is 0.18 times the number of pencils)
To make a box and whisker plot, you need to draw a number line that would be able to represent all the values in the data (look at your lowest and highest numbers).
You will need to graph the following 5 points on your number line (median, lower quartile, lower extreme, upper quartile, and upper extreme).
Create a box around the quartiles and extend the lines to the extremes.
See the picture I have attached for the box and whisker box for this data set.