The zeros are x= 0,-5,-8. the zeros are the x values where the graph intersects the x-axis. to find the zeros, you have to replace y with zero then solve for x.
Answer:
15
Step-by-step explanation:
take the triangle away from the square you will get 9 for the square 3x3 you get the 3 from 7 because 7-3=4 the 4 if for the triangle 4x3=12 divided by 2 bc its a triangle u get 6 6+9=15
<h3>
Answer: -13</h3>
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Explanation:
g(-3) = 2 means x = -3 and y = 2 pair up together to form the point (-3,2)
g(1) = -4 means we have the point (1,-4)
Find the slope of the line through the two points (-3,2) and (1,-4)
m = (y2-y1)/(x2-x1)
m = (-4-2)/(1-(-3))
m = (-4-2)/(1+3)
m = -6/4
m = -3/2
m = -1.5
The general slope intercept form y = mx+b turns into y = -1.5x+b after replacing m with -1.5
Plug in (x,y) = (-3,2) which is one of the points mentioned earlier and we end up with this new equation: 2 = -1.5*(-3) + b
Let's solve for b
2 = -1.5*(-3)+b
2 = 4.5 + b
2-4.5 = 4.5+b-4.5 .... subtract 4.5 from both sides
-2.5 = b
b = -2.5
Therefore, y = mx+b becomes y = -1.5x-2.5 meaning the g(x) function is g(x) = -1.5x-2.5
The last step is to plug in x = 7 and compute
g(x) = -1.5*x - 2.5
g(7) = -1.5*7 - 2.5
g(7) = -10.5 - 2.5
g(7) = -13
Answer:
Explanation:
<u />
<u>1. First find the density of your chain</u>
- Volume = displaced water volume
= Volume of Final level of water - initial level of water
= 20 ml - 15 ml = 5 ml
- Density = 66.7g / 5 ml = 13.34 g/ml
<u />
<u>2. Second, write the denisty of the chain as the weighted average of the densities of the other metals:</u>
Mass of gold × density of gold + mass of other metals × density of other metals, all divided by the mass of the chain.
Calling x the amount of gold, then the amount of other metals is 66.7 - x:
Then, there are 26.47 grams of gold in 66.7 grams of chain, which yields a percentage of:
- (26.47 / 66.7) × 100 = 39.7%
Answer:
Step-by-step explanation:
Let x represent the number of years it will take the two colleges to have the same enrollment.
In 2000, there were 12900 students at college A, with a projected enrollment increase of 900 students per year. This means that the expected number of students at college A in x years time is
12900 + 900x
In the same year, there were 25,000 students at college B, with a projected enrollment decline of 700 students per year. This means that the expected number of students at college B in x years time is
25000 - 700x
For both colleges to have the same enrollment,
12900 + 900x = 25000 - 700x
900x + 700x = 25000 - 12900
1600x = 12100
x = 12100/1600
x = 7.56
Approximately 8 years
The year would be 2000 + 8 = 2008
