Ask question

Ask question

All categories

3 years ago

12

A table of values of a linear function is shown below. Find the output when the input is . Type your answer in the space provide

d.
input 1234n
output6420

1 answer:

Tcecarenko [31]3 years ago

6 0

Answer:

y=4n+2

really sorry if this aint right

Step-by-step explanation:

You might be interested in

Can any one explain how to look for d direction of a vector .

Shkiper50 [21]

Answer:

The direction of a vector is the measure of the angle it makes with a horizontal line . tanθ=y2 − y1x2 − x1 , where (x1,y1) is the initial point and (x2,y2) is the terminal point. Example 2: Find the direction of the vector →PQ whose initial point P is at (2,3) and end point is at Q is at (5,8)

7 0

3 years ago

The table below represents a linear function. What is the slope?

BartSMP [9]

It’s d because if you reverse -2 to go up the graph instead of down you get 0, then 2, then 4

4 0

2 years ago

Explain how solve 4^(x+3) = 7 using the change of base formula log base b of y equals log y over log b. Include the solution for

MaRussiya [10]

Answer:

-1.596

Step-by-step explanation:

Apply log base 4 both sides

log4(4^(x+3)) = log4(7)

(x+3)log4(4) = lg7/lg4

x+3 = 1.403677461

x = -1.596322539

4 0

3 years ago

If y = -1.02 when x= 5.1, what is y when x = 7.1?

MakcuM [25]

Answer:

y = -1.42

Step-by-step explanation:

(7.1/5.1) × -1.02 = -1.42

5 0

3 years ago

What is the greatest value in the range of f(x) = x^2 - 3 for the domain (-3,0,1,2)

bezimeni [28]

Answer:

f(-3)=6 is the greatest value in the range of $f(x)=x^2-3$ for the domain (-3,0,1,2)

Step-by-step explanation:

Given that the function f is defined for range by $f(x)=x^2-3$ for the domain (-3,0,1,2)

To find the greatest value in the range of $f(x)=x^2-3$ for the domain (-3,0,1,2):

$f(x)=x^2-3$ for the domain (-3,0,1,2)

That is put x=-3 in the given function $f(x)=x^2-3$ we get

$f(-3)=(-3)^2-3$

$=3^2-3$

$=9-3$

$=6$

Therefore f(-3)=6

put x=0 in the given function $f(x)=x^2-3$ we get

$f(0)=(0)^2-3$

$=0-3$

$=-3$

Therefore f(0)=-3

put x=1 in the given function $f(x)=x^2-3$ we get

$f(1)=(1)^2-3$

$=1-3$

$=-2$

Therefore f(1)=-2

put x=-3 in the given function $f(x)=x^2-3$ we get

$f(2)=(2)^2-3$

$=4-3$

$=1$

Therefore f(2)=1

Comparing the values of f(-3)=6,f(0)=-3,f(1)=-2,and f(2)=1 to find the greatest value in the range of f(x) = x^2 - 3 for the domain (-3,0,1,2) we get

Therefore f(-3)=6 is the greatest value in the range of $f(x)=x^2-3$ for the domain (-3,0,1,2)

8 0

4 years ago