Answer:
B
Step-by-step explanation:
So far, we know that:
∠D = ∠J.
And that:
DE:JK = 14:7 = 2:1
So, to prove that ΔDEF ~ ΔJKL by SAS, DF must be similar to JL, as those are the sides between the angle.
So:
DF:JL = 2:1.
Our answer is B.
Part 1: Answer:
(x+1)(x+1)(x-6) = x^3 - 4x^2 - 11x - 6
Step-by-step explanation:
To make r a root, include (x-r) as a factor. (-1+1)(-1+1)(-1-6) is zero even though (-1-6) isn't.
(6+1)(6+1)(6-6) is zero.
Part 2 Answer:
Standard form: y = -x^4 + 12
Degree 4
left end goes down, right end goes down.
Step by step: apply the definitions of standard form, polynomial degree, and "end behavior". In other words, read the textbook.
Part 3: Answer: x = 3, x = 8
Step by step:
x^2-11x = -24
x^2-11x+24 = 0
(x-3)(x-8) = 0
x = 3 or x = 8
Part 4a Answer:
quotient 2x^2 + x - 3
remainder 1
Step by step:
2x^2 + x - 3
___________________
x-4 ) 2x^3 - 7x^2 - 7x + 13
2x^3 - 8x^2
__________
0 + x^2 - 7x + 13
x^2 - 4x
____________
0 - 3x + 13
- 3x + 12
______
1
Part 4b answer:
quotient 2x^2 - 6x + 2
remainder -20
Step by step: you have to know exactly what you are doing. Refer to textbook or Wikipedia.
dividend 2x^3 +14x^2 - 58x
divisor x+10
leading coefficient of divisor must be 1
write coefficients of dividend at top
write coefficients of dividend at left
| 2 14 -58 0
-10 | -20 60 -20
___________
| 2 -6 2 -20
Coefficients of quotient are 2 -6 2
Remainder is -20
quotient = 2x^2 - 6x + 2
I see. Imagine you have f(x)=|x|. It's a V shaped graph.
Now if f(x)=|x|, 2f(x)=2|x|.
Graph Transformation Rule:
af(x), multiply y-coordinates by a.
*Ultimately, you'd still have a V shaped graph in 2f(x)=2|x|, but the y values of all the coordinates in f(x)=|x| would have to be multiplied by 2 giving you 2f(x)=2|x|.
Answer:
area formula: base x height 9x4=36
Step-by-step explanation:
The y-intercept represents the flat charge to hire the plumber
Explanation: The y-intercept is 0 hour, which can represent flat charge