Answer:
<3 and <6 is Alternate interior angles.
<4 and <5 is Alternate interior angles.
<2 and <7 is Alternate exterior angles.
<3 and <7 is Corresponding angles.
Hope this helps. :)
41/6 = - 41/6 = - 6 5/6 = - 6.833333333333; Result written as a negative improper fraction, a mixed number (also called a mixed fraction) and a decimal number. Ordinary (simple, common) math fraction reducing to lowest terms (simplifying), answer with explanations below.
Answer:
m=(-6,6)
Step-by-step explanation:
m^2 -36 = 0
Reorder the terms:
-36 + m^2 = 0
Solving for variable 'm'.
Add '36' to each side of the equation.
-36 + 36 + m^2 = 0 + 36
Combine like terms: -36 + 36 = 0
0 + m^2 = 0 + 36
m^2 = 0 + 36
Combine like terms: 0 + 36 = 36
m^2 = 36
Simplifying
m^2 = 36
Take the square root of each side:
√m^2=+/-√36
m=(+/-)6
m = {-6, 6}
Answer: The value of
is 36.
Step-by-step explanation:
Given expression: ![6(2b-4)](https://tex.z-dn.net/?f=6%282b-4%29)
To find the value of
at b= 5, we need to substitute the b=5 in the expression, we get
![6(2(5)-4)\\=6(10-4).......[\text{solve parentheses}]\\=6(6)\\=6\times6=36](https://tex.z-dn.net/?f=6%282%285%29-4%29%5C%5C%3D6%2810-4%29.......%5B%5Ctext%7Bsolve%20parentheses%7D%5D%5C%5C%3D6%286%29%5C%5C%3D6%5Ctimes6%3D36)
![\\\Rightarrow6(2(5)-4)=36](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow6%282%285%29-4%29%3D36)
Therefore, the value of
is 36, when b=5.
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.