Answer:
m = 7.1
Step-by-step explanation:
3 (m -5) + 8 = 7 -1/2 (4m - 11)
So, what I did first was distribute the values in the parenthesis to get,
3m - 15 + 8 = 7 -2m -5.5
Now that we have the parenthesis taken care of we can do the simpler math,
3m - 23 = 7 -2m - 5.5
I just added the 15 and 8, so now I move the -5.5 to the opposite side by adding.
3m - 23 = 7 -2m -5.5
<u> +5.5 +5.5</u>
3m - 28.5 = 7 - 2m
Here we can do the same with the -2m.
3m - 28.5 = 7 - 2m
<u>+2m +2m</u>
5m - 28.5 = 7
To get rid of the -28.5 I added it to the 7 getting an answer of,
5m - 28.5 = 7
<u> + 28.5 +28.5</u>
5m = 35.5
Finally, divide 5m and 35.5 both by 5.
5m/5 = m
35.5/5 = 7.1
Answer: m=7.1
Sorry it's really long, but I hope this helps! Have a great day!
The value of f(a)=4-2a+6
, f(a+h) is
, [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6
.
Given a function f(x)=4-2x+6
.
We are told to find out the value of f(a), f(a+h) and [f(a+h)-f(a)]/hwhere h≠0.
Function is like a relationship between two or more variables expressed in equal to form.The value which we entered in the function is known as domain and the value which we get after entering the values is known as codomain or range.
f(a)=4-2a+6
(By just putting x=a).
f(a+h)==
=4-2a-2h+6(
)
=4-2a-2h+6
=
[f(a+h)-f(a)]/h=[
-(4-2a+6
)]/h
=
=
=6h+12a-2.
Hence the value of function f(a)=4-2a+6
, f(a+h) is
, [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6
.
Learn more about function at brainly.com/question/10439235
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Answer:
Distance LM = 5.20 unit (Approx.)
Step-by-step explanation:
Given coordinates;
L(1, 4, 7) and M(2, 9, 8)
Find:
Distance LM
Computation:
Distance between three-dimensional plane = √(x2 - x1)² + (y2 - y1)² + (z2 - z1)²
Distance LM = √(2 - 1)² + (9 - 4)² + (8 - 7)²
Distance LM = √(1)² + (5)² + (1)²
Distance LM = √1 + 25 + 1
Distance LM = √27
Distance LM = 3√3 unit
Distance LM = 3(1.732)
Distance LM = 5.196
Distance LM = 5.20 unit (Approx.)
Answer:
Here you go
Step-by-step explanation:
x=0,-4
Answer:
A Euler circuit has TWO ODD vertices
Step-by-step explanation: