Answer:
f + g)(x) = f (x) + g(x)
= [3x + 2] + [4 – 5x]
= 3x + 2 + 4 – 5x
= 3x – 5x + 2 + 4
= –2x + 6
(f – g)(x) = f (x) – g(x)
= [3x + 2] – [4 – 5x]
= 3x + 2 – 4 + 5x
= 3x + 5x + 2 – 4
= 8x – 2
(f × g)(x) = [f (x)][g(x)]
= (3x + 2)(4 – 5x)
= 12x + 8 – 15x2 – 10x
= –15x2 + 2x + 8
\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(
g
f
)(x)=
g(x)
f(x)
= \small{\dfrac{3x+2}{4-5x}}=
4−5x
3x+2
My answer is the neat listing of each of my results, clearly labelled as to which is which.
( f + g ) (x) = –2x + 6
( f – g ) (x) = 8x – 2
( f × g ) (x) = –15x2 + 2x + 8
\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}
Answer:
There are two types of similar triangle problems; these are problems that require you to prove whether a given set of triangles are similar and those that require you to calculate the missing angles and side lengths of similar triangles. Subtract both sides by 130°. Hence; By Angle-Angle (AA) rule, ΔPQR~ΔXYZ.
Step-by-step explanation:
5y - 2x + 1 = 0.
5y = 2x-1.
y = 2/5 x - 2/5
this tells us that tan(theta) = 2/5
sin(theta)/sqrt(1-sin^2(theta)) = 2/5
x/sqrt(1-x^2)=2/5
x^2/(1-x^2)=4/25
25x^2=4-4x^2
29x^2=4
x^2=4/29
x = -2/sqrt(29) because quadrant III
cos(x) = sqrt(1-4/29) = -5/sqrt(29)
tan(theta)=2/5
cos(theta)=-5/sqrt(29)
sin(theta)=-2/sqrt(29)
cot(theta)=5/2
sec(theta)=-sqrt(29)/5
csc(theta)=-sqrt(29)/2
Answer:
-8
Step-by-step explanation:
you have to make them equivalent so and the one thing missing from the other side of the equation
1:24 get them each doqwn 1 inch 24 inche = 2 feet
