If f(x)=-x+8 and g(x)= x^4 what is (g°f)(2)?

1 answer:

3 0

Notice, is really the same. All we do is, "insert one function into the other".

so, a way to do it is, get the value of the first one, then insert that output into the second.

$\bf \begin{cases}
f(x)=-x+8\\
g(x)=x^4\\
(g\circ f)(x)=g(~~f(x)~~)
\end{cases}
\\\\\\
(g\circ f)(2)=g(~~f(2)~~)
\\\\\\
f(2)=-(2)+8\implies \boxed{f(2)=6}
\\\\\\
g(~~f(2)~~)\implies g\left(~~\boxed{6}~~ \right)\implies g(6)=(6)^4\implies g(6)=1296$

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3 ⅕ -⅔ divided by 2 ⅘

spin [16.1K]

Answer:

160/196

Step-by-step explanation:

it would be easier if you converted all fractions to improper fractions

so 3 1/5 would equal 16/5 and 2 4/5 would equal to 14/5

$\frac{14}{5} - \frac{2}{3}$ : to solve this you would need the same common denomenator,

to do this you can multiply 5 and 3 by something to make the lowest common denominator, which would be 15

thus

multiplying first fraction by 3 and the second by 5

42/15 - 10/15 = 32/15

now 32/15 divided by 14/5

the rule for dividing fractions, is flipping the fraction which you will be using to divide, and then multiplying it to the first fraction:

$\frac{32}{14}$ multiplied by $\frac{5}{14}$

thus, answer is 160/196

8 0

3 years ago

The coordinate of the vertices of parallelogram CDEH are C(-5,5), D(2,5), E(-1,-1), and H(-8,-1). What are the coordinates of P,

d1i1m1o1n [39]

Given:

Vertices of a parallelogram are C(-5,5), D(2,5), E(-1,-1), and H(-8,-1).

P is the intersection point of diagonals CE and DH.

To find:

The coordinates of P.

Solution:

We know that, diagonals of a parallelogram always bisect each other. It means the intersection point of the diagonal is the midpoint point of both diagonals.

We can find the midpoint of either diagonal CE or diagonal DH to get the coordinates of intersection point of diagonals, i.e. P.

So, point P is midpoint of CE. So,

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$