Answer: The first day the author reaches 100 days is on day 16.
To solve this problem, you could use a graphing calculator to graph the given equation. Then, determine when this line crosses 100. It crosses when x = 15.539. Therefore, we would have to round up to 16 so it is at least 100.
You could use the quadratic equation to solve: 100 = x^2 -12x + 45
Either you will get 16. If you use the quadratic formula, make sure to only use the positive answer.
Answer:
15 kilometers
Step-by-step explanation:
So we know:
For 4 days she runs 1500 meters each day.
For 3 days she runs 3 kilometers each day.
First off, lets convert 1500 meters into kilometers.
There are 1000 meters in a kilometer, so pluggin in 1500 meters:
1500/1000 = 1.5
So for 4 days she runs 1.5 kilometers each day.
Now, to find the total kilometers, we must find multiply the 1.5 kilometers by the 4 days, the 3 kilometers by the 3 days, then add those two together.
So lets do this:
1.5*4 = 6
So she ran 6 kilometers in the first 4 days.
Next we have:
3*3 = 9
So she ran 9 kilometers in the last 3 days.
Now finally add them together:
6 + 9 = 15
So she ran a total of 15 kilometers.
Hope this helps!
Answer:
solution set is (x,y) = (7,6)
Step-by-step explanation:
solving by substitution method
2x +y=20--------------1
6x-5y=12---------------2
from equation 1, solve for y
2x+y=20
y= 20-2x------equation 3
adding value of y in equation 2
6x-5y=12
6x-5(20-2x)=12
6x-100+10x=12
16x= 12+100
16x= 112
x= 112/16
x=7
adding value of x in equation 3
y= 20-2x
y= 20- 2(7)
y=20-14
y=6
so solution set (x,y) = (7,6)
<h3>
Answer: False</h3>
==============================================
Explanation:
I'm assuming you meant to type out
(y-2)^2 = y^2-6y+4
This equation is not true for all real numbers because the left hand side expands out like so
(y-2)^2
(y-2)(y-2)
x(y-2) .... let x = y-2
xy-2x
y(x)-2(x)
y(y-2)-2(y-2) ... replace x with y-2
y^2-2y-2y+4
y^2-4y+4
So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4
--------------------------
Another approach is to pick some y value such as y = 2 to find that
(y-2)^2 = y^2-6y+4
(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2
0^2 = 2^2 - 6(2) + 4
0 = 4 - 6(2) + 4
0 = 4 - 12 + 4
0 = -4
We get a false statement. This is one counterexample showing the given equation is not true for all values of y.
Answer:
42.43 ft
Step-by-step explanation:
Please find attached an image of the softball pitch
h² = p² + f² -2x p×f×cos45
h² = 43² + 60² - 2x (43)(60) x 0.7071
h² = 1849 + 3600 - 3648.636
h² = 1800.364
h = √1800.364
h = 42.43 feet
