Answer:
y = 3x +4
Step-by-step explanation:
The equation for the parallel line will have the same x- and y-coefficients, but a different constant. You can put the given point values into the equation to see what the constant needs to be:
y = 3x + b
10 = 3·2 + b . . . . . . substitute x=2, y=10
4 = b . . . . . . . . . . . . subtract 6
The equation of the line is ...
y = 3x +4
Answer: ![3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Work Shown:
![\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27x%5E%7B6%7Dy%5E%7B4%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%2B3%7Dy%5E%7B3%2B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%7D%2Ax%5E%7B3%7D%2Ay%5E%7B3%7D%2Ay%5E%7B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B2%2A3%7D%2Ay%5E%7B3%7D%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%7D%2A%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Explanation:
As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.
What is 19 and 20 you didn’t put anything
Answer:
1/sqrt10
Step-by-step explanation:
1) Find out cosA using formula (cosA)^2+(sinA)^2=1
The module of cosA= sqrt (1- (-3/5)^2)= sqrt 16/25=4/5
So cosA=-4/5 or cosA=4/5.
Due to the condition 270degrees< A<360 degrees, 0<cosA<1 that's why cosA=4/5.
2) Find sinA/2 using a formula cosA= 1-2sinA/2*sinA/2 where cosA=4/5.
(sinA/2)^2= 0.1
sinA= sqrt 0.1= 1/ sqrt10 or sinA= - sqrt 0.1= -1/sqrt10
But 270°< A< 360°, then 270/2°<A/2<360/2°
135°<A/2<180°, so sinA/2 must be positive and the only correct answer is
sin A/2= 1/sqrt10
