-1+3m=2m+4
put the like terms together
3m-2m=4+1
m=5
Distribute 2/5 and 3/5 into the ():
2/5(a+b)+3/5(a+c)
2/5 a+ 2/5 b+3/5 a+ 3/5 c
combine the like terms:
2/5 a+3/5 a= 5/5 a --> 1a --> a
new simplified equation:
a+2/5 b+3/5 c
So lets try to prove it,
So let's consider the function f(x) = x^2.
Since f(x) is a polynomial, then it is continuous on the interval (- infinity, + infinity).
Using the Intermediate Value Theorem,
it would be enough to show that at some point a f(x) is less than 2 and at some point b f(x) is greater than 2. For example, let a = 0 and b = 3.
Therefore, f(0) = 0, which is less than 2, and f(3) = 9, which is greater than 2. Applying IVT to f(x) = x^2 on the interval [0,3}.
Learn more about Intermediate Value Theorem on:
brainly.com/question/11377865
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Answer:
b(-10) = 6
Step-by-step explanation:
Step 1: Define
b(x) = |x + 4|
b(-10) is x = -10
Step 2: Substitute and Evaluate
b(-10) = |-10 + 4|
b(-10) = |-6|
b(-10) = 6
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