Answer:
Step-by-step explanation:
Are you in Calculus? These are calculus concepts!
To calculate the rate of change here you must specify an interval, e. g., "what is the rate of change on the interval (0, 3)?"
If you know calculus: The 'rate of change' on the interval (a, b) is
f(b) - f(a)
r. of c. = --------------------
b - a
Have you used this formula before?
Because of the 'x^2' term this is NOT a linear function.
If you want more explanation, provide an interval on which you want the average rate of change and ask specific questions of your own.
Hi again
x² - 2x - 8
Factor
= (x + 2)(x -4)
Answer : A is the correct option.
If you wanna check if A is the right answer, I will solve it for you and it will you give you the exact answer which will be x² - 2x - 8
(x+2)(x-4)
Distribute
(x)(x) + (x)(-4)+(2)(x)+(2)(-4)
= x² - 4x + 2x - 8
= x² - 2x - 8
Here it is, we find the exam answer.
Thus, (x+2)(x-4) is the reciprocal of x² -2x - 8
I hope that's help !
Answer:
PROOF FOR THE "PROVE" SECTION:
As linear pairs, angle 2 and 3 are supplementary to each other. Angle 1 is equal to angle 2, as they are both same-side interior angles. Therefore, angle 1 and angle 3 are also supplementary.
Filling in the missing blanks:
S1. Angle 1, Angle 2, Angle 3
S2. Angle 1 and Angle 2
R3. Congruent (___)
R5. supplementary angles
S7. Angle 1 = Angle 2, so Angle 1 can be substitued in for Angle 2 in any equation, and Angle 2 can be substitued for Angle 1 in any equation as well (they can replace each other, like x=y & y=x or a=b & b=a)
Hope this helped! Have a great day (pls mark brainliest)!!
Answer:
Answer for 2nd is option c, for 3rd is option d, for 4th is option e
Step-by-step explanation:
As we know 1 ft.=12 in.
- In ΔABC
∴ The congruent sides are AB and AC respectively
- CB =12 ft. 4 in.=148 in.
- AB=
CB =111 in. =9 ft. 3 in. - AC=CB =111 in. =9 ft. 3 in.
∵ <em>Perimeter of ΔABC</em> =AB+AC+CB
=9 ft. 3 in. + 9 ft. 3 in. +12 ft. 4 in.
=30 ft. 10 in.
2. In ΔDEF
∴ The congruent sides are DE and DF respectively
- DE = 6 ft. 3 in. =75 in.
- DF = 6 ft. 3 in. =75 in.
- Let the length of FE is equal to x
- 0.75FE =DE =DF
- 0.75x = 6 ft. 3 in. =75 in.
- x =100 in. =8 ft. 4 in.
∵ <em>Perimeter of ΔDEF</em> =DE+DF+FE
= 6 ft. 3 in. +6 ft. 3 in. +8 ft. 4 in.
= 20 ft. 10 in.
3. In ΔJKL
∴ The congruent sides are JL and KL respectively
- JK = x+3
- KL =4x-17
- JL =6x-45
- JL≅KL
- 4x-17 =6x-45 . . . . . . . . . . . . . . . . . . . . . . . (1)
- Subracting 4x from both sides from eq 1
- -17 =2x-45
- Adding 45 on both the sides
- 28 =2x
- Dividing by 2 on both sides
- 14 =x
- JK = 14+3 =17
- KL = 4×14-17 =39
- JL = 6×14-45 =39
∵ <em>The dimensions of the ΔJKL are 39,39 and 17.</em>
