-x<-x+7(x-2)
-x<-x+7x-14 Distribute the 7
-x<6x-14 Combine like terms
-7x<-14 Move all variables to one side
x>2 Divide by -7 to isolate the variable
Answer:
B. False
Step-by-step explanation:
There is not enough information to make that conclusion. The two statements are completely unrelated, so the transitive property cannot be used. None of the given statements say that ABC is congruent to MNO or PQR. That means that nothing can be assumed about DEF. To use the transitive property you would need proof that ABC=MNO or ABC=PQR. But neither of those statements are there so the answer is false.
Answer:
Please check the wording of the question. There isn't enough information to select a single value of x.
Step-by-step explanation:
The attached worksheet shows two different combinations of pens and pencils that both sum to a total cost of $10. There are many other combinations that would do the same.
Answer:
Step-by-step explanation:
3x² + 5x + 2 = 3x² + 3x + 2x + 2
=3x*(x + 1) + 2*(x + 1)
= (x + 1)(3x + 2)
So it tells us that g(3) = -5 and g'(x) = x^2 + 7.
So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))
now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16
Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84
So we have our linear approximation for the two.
If you wanted to, you could check your answer by finding g(x). Since you know g'(x), take the antiderivative and we will get
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35
So just to check our linear approximations use that to find g(2.99) and g(3.01)
g(2.99) = -5.1597
g(3.01) = -4.8397
So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer. Not a bad method if you ever need to use it.
