It seems that some the work is already here, but I'd be glad to!! So for #3 which is 9x^2+15x, we can factor out both a 3 and an x (3x) so we know that 3x * 3x =9x^2 and 3x * 5 = 15x so once we take the 3x out of the equation, we are left with 3x(3x+5) and that's as far as you can factor.
For #4, we see that the common factor is 10m because 10m * 2n = 20mn and 10m * 3 = 30m so once we take 10m out of the original, it becomes 10m(2n-3)
For #5, this one the common factor is 4xy because 4xy * 2xy=8x^2y^2 and 4xy*x= 4x^2y and 4xy*3=12xy so once we take the 4xy out of the equation, it becomes 4xy(2xy-x-3)
Hope this helps!
Answer:
The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.
Step-by-step explanation:
Given information:
A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.
![z=1](https://tex.z-dn.net/?f=z%3D1)
![\frac{dx}{dt}=430](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%3D430)
We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.
![y=2](https://tex.z-dn.net/?f=y%3D2)
According to Pythagoras
![hypotenuse^2=base^2+perpendicular^2](https://tex.z-dn.net/?f=hypotenuse%5E2%3Dbase%5E2%2Bperpendicular%5E2)
![y^2=x^2+1^2](https://tex.z-dn.net/?f=y%5E2%3Dx%5E2%2B1%5E2)
.... (1)
Put z=1 and y=2, to find the value of x.
![2^2=x^2+1^2](https://tex.z-dn.net/?f=2%5E2%3Dx%5E2%2B1%5E2)
![4=x^2+1](https://tex.z-dn.net/?f=4%3Dx%5E2%2B1)
![4-1=x^2](https://tex.z-dn.net/?f=4-1%3Dx%5E2)
![3=x^2](https://tex.z-dn.net/?f=3%3Dx%5E2)
Taking square root both sides.
![\sqrt{3}=x](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%3Dx)
Differentiate equation (1) with respect to t.
![2y\frac{dy}{dt}=2x\frac{dx}{dt}+0](https://tex.z-dn.net/?f=2y%5Cfrac%7Bdy%7D%7Bdt%7D%3D2x%5Cfrac%7Bdx%7D%7Bdt%7D%2B0)
Divide both sides by 2.
![y\frac{dy}{dt}=x\frac{dx}{dt}](https://tex.z-dn.net/?f=y%5Cfrac%7Bdy%7D%7Bdt%7D%3Dx%5Cfrac%7Bdx%7D%7Bdt%7D)
Put
, y=2,
in the above equation.
![2\frac{dy}{dt}=\sqrt{3}(430)](https://tex.z-dn.net/?f=2%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Csqrt%7B3%7D%28430%29)
Divide both sides by 2.
![\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Cfrac%7B%5Csqrt%7B3%7D%28430%29%7D%7B2%7D)
![\frac{dy}{dt}=372.390923627](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D372.390923627)
![\frac{dy}{dt}\approx 372](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%5Capprox%20372)
Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.
B. 250 cm3 hope this helps
Answer: True becuase 5+8 equals 13 and 6+7 also equal 13 so that means that they are equal to each other.
Step-by-step explanation:
Answer:
It provides different and sometimes opposing viewpoints about an event.