Answer:
x = - 5 , x = 
Step-by-step explanation:
the values of x that make f(x) zero are the zeros
to find the zeros let f(x) = 0 , that is
3x² + 13x - 10 = 0
consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 3 × - 10 = - 30 and sum = + 13
the factors are + 15 and - 2
use these factors to split the x- term
3x² + 15x - 2x - 10 = 0 ( factor the first/second and third/fourth terms )
3x(x + 5) - 2(x + 5) = 0 ← factor out (x + 5) from each term
(x + 5)(3x - 2) = 0
equate each factor to zero and solve for x
x + 5 = 0 ⇒ x = - 5
3x - 2 = 0 ⇒ 3x = 2 ⇒ x = 
Answer:
let the number be x
4x-6=3x+8
Step-by-step explanation:
hope this helps you
If yes then do mark my answer as brainliest
Answer: (f-g)(2)=14
Step-by-step explanation:
(f – g) (-2) means the same as subtracting f(2) and g(2). Since we are given f(x) and g(x), we can use them to solve. There are two ways to solve. One is to find f(2) and g(2), and then subtract them. Another way is to do (f-g)(x), then plug in x=2. I will show both methods.
Method 1
f(2)=3(2)²+1 [exponent]
f(2)=3(4)+1 [multiply]
f(2)=12+1 [add]
f(2)=13
g(2)=1-(2) [subtract]
g(2)=-1
(f-g)(2)=13-(-1) [subtract f(2) and g(2)]
(f-g)(2)=14
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Method 2
(f-g)(x)=3x²+1-(1-x) [distribute -1]
(f-g)(x)=3x²+1-1+x [combine like terms]
(f-g)(x)=3x²+x
(f-g)(2)=3(2)²+2 [plug in x=2, exponent]
(f-g)(2)=3(4)+2 [multiply]
(f-g)(2)=12+2 [add]
(f-g)(2)=14
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Now, we know that (f-g)(2)=14. We confirmed this with both methods.
Answer:
The graph of the circle is not a function
Answer:
C) Tom did not distribute to both terms in parentheses.
Step-by-step explanation:
Addition within a paranthesis has a distributive property to the multiplier outside the paranthesis. Ignoring this will lead to a wrong value for the operation.