Answer:
y = 8x + 38
Step-by-step explanation:
To find the slope we set up y = mx + b and plug m (our slope) in and then plug in the point into x and y to find b (our y-intercept)
y = mx + b
y = 8x + b
6 = 8(-4) + b
6 = -32 + b
+32 +32
38 = b
So using our slope and our intercept we have our slope intercept
y = mx + b
y = 8x + 38
Hope this helps :)
625,087,435
600,000,000 + 20,000,000 + 5,000,000 + 80,000 + 7,000 + 400 + 30 + 5
Answer:
182 seconds
Step-by-step explanation:
Given:
Time taken for first lap is 50 seconds and each subsequent lap takes 20% longer than previous
Solution:
Time taken for first lap = 50 sec
For second lap
Each subsequent lap takes 20% longer than previous
Previous lap time = 50 sec
second lap = 50 + 20% of 50
second lap ![=50+(\frac{20}{100}\times 50)](https://tex.z-dn.net/?f=%3D50%2B%28%5Cfrac%7B20%7D%7B100%7D%5Ctimes%2050%29)
second lap ![= 50 + 10](https://tex.z-dn.net/?f=%3D%2050%20%2B%2010)
second lap ![= 60\ sec](https://tex.z-dn.net/?f=%3D%2060%5C%20sec)
For third lap
Each subsequent lap takes 20% longer than previous
Previous lap time = 60 sec
Third lap = 60 + 20% of 60
![Third\ lap = 60 + (\frac{20}{100}\times 60)](https://tex.z-dn.net/?f=Third%5C%20lap%20%3D%2060%20%2B%20%28%5Cfrac%7B20%7D%7B100%7D%5Ctimes%2060%29)
![Third\ lap = 60+2\times 6](https://tex.z-dn.net/?f=Third%5C%20lap%20%3D%2060%2B2%5Ctimes%206)
![Third\ lap = 60+12](https://tex.z-dn.net/?f=Third%5C%20lap%20%3D%2060%2B12)
![Third\ lap = 72\ sec](https://tex.z-dn.net/?f=Third%5C%20lap%20%3D%2072%5C%20sec)
Time taken for 3 laps is equal to sum of all three laps time.
![Tiem\ taken\ for\ 3\ lap = 50+60+72](https://tex.z-dn.net/?f=Tiem%5C%20taken%5C%20for%5C%203%5C%20lap%20%3D%2050%2B60%2B72)
![Tiem\ taken\ for\ 3\ lap = 182\ sec](https://tex.z-dn.net/?f=Tiem%5C%20taken%5C%20for%5C%203%5C%20lap%20%3D%20182%5C%20sec)
Therefore, it takes 182 seconds (3 min, 2 seconds) to run three laps
Answer:
Yes
Step-by-step explanation:
Since the longest side is 6.5 cm, so let it be hypotenuse c.
Verifying by right angle formula
c^2=a^2+b^2
(6.5)^2=(3.9)^2+(5.2)^2
42.25=15.21+27.04
42.25=42.25
Hence the given triangle is a right angles triangle
Answer:
The dolphin will be above the surface of the water for 2 seconds.
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
.
This polynomial has roots
such that
, given by the following formulas:
![x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D%20%5Cfrac%7B-b%20%2B%20%5Csqrt%7B%5CDelta%7D%7D%7B2%2Aa%7D)
![x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%20%5Cfrac%7B-b%20-%20%5Csqrt%7B%5CDelta%7D%7D%7B2%2Aa%7D)
![\Delta = b^{2} - 4ac](https://tex.z-dn.net/?f=%5CDelta%20%3D%20b%5E%7B2%7D%20-%204ac)
The height of the dolphin after t seconds is given by:
![h(t) = -16t^2 + 96t - 128](https://tex.z-dn.net/?f=h%28t%29%20%3D%20-16t%5E2%20%2B%2096t%20-%20128)
According to Micha's model, how long will the dolphin be above the surface of the water?
It stays above the surface of the water between the first and the second root. Initially, it is below water, when the first time for which
it crosses the surface upwards, and then the second time for which
it crosses the surface downwards.
We have to find these roots. So
![h(t) = -16t^2 + 96t - 128](https://tex.z-dn.net/?f=h%28t%29%20%3D%20-16t%5E2%20%2B%2096t%20-%20128)
![-16t^2 + 96t - 128 = 0](https://tex.z-dn.net/?f=-16t%5E2%20%2B%2096t%20-%20128%20%3D%200)
Multiplying by -16
![t^2 - 6t + 8 = 0](https://tex.z-dn.net/?f=t%5E2%20-%206t%20%2B%208%20%3D%200)
![\Delta = (-6)^{2} - 4*1*8 = 36 - 32 = 4](https://tex.z-dn.net/?f=%5CDelta%20%3D%20%28-6%29%5E%7B2%7D%20-%204%2A1%2A8%20%3D%2036%20-%2032%20%3D%204)
![t_{1} = \frac{-(-6) + \sqrt{4}}{2} = 4](https://tex.z-dn.net/?f=t_%7B1%7D%20%3D%20%5Cfrac%7B-%28-6%29%20%2B%20%5Csqrt%7B4%7D%7D%7B2%7D%20%3D%204)
![t_{2} = \frac{-(-6) - \sqrt{4}}{2} = 2](https://tex.z-dn.net/?f=t_%7B2%7D%20%3D%20%5Cfrac%7B-%28-6%29%20-%20%5Csqrt%7B4%7D%7D%7B2%7D%20%3D%202)
4 - 2 = 2
The dolphin will be above the surface of the water for 2 seconds.