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3 years ago

Which table shows a linear fucntion

Mathematics

2 answers:

exis [7]3 years ago

8 0

Plz show us the table so I can help you

Katena32 [7]3 years ago

3 0

Answer:

can you attach what your talking about?

Step-by-step explanation:

You might be interested in

3. Find the range of this function.

valkas [14]

Answer:

The answer is three

Step-by-step explanation:

7 0

3 years ago

1= 1/4c what is c ? pls help i could not figure this one out haha

tino4ka555 [31]

Step-by-step explanation:

it say the answer is 4 but I don't know how

8 0

3 years ago

Claire Drew model of 5/10 explain how Claire could change her model to show how many hundreds are in 5/10 I know this is like my

natta225 [31]

Answer:

There are 0.005 hundreds in 5/10.

Step-by-step explanation:

Claire drew model of 5/10

We want to know how many hundreds are in 5/10.

Let us use an obvious example.

There are three 2's in 6 right?

Suppose we didn't know this, and we are told to find how many 2's are in 6, we get this by representing this in an algebraic expression as:

There are x 2's in 6. This can be written as

2x = 6

Solving for x, by dividing both sides by 2, we have the number of 2's that are in 6.

x = 6/2 = 3.

Now, to our work

We want to find how many hundreds are in 5/10. We solve the equation

100x = 5/10

x = 5/1000 = 0.005

There are 0.005 hundreds in 5/10.

7 0

3 years ago

A basic cellular phone plan costs $4 per month for 70 calling minutes. Additional time costs $0.10 per minute. The formula C= 4+

Harrizon [31]

Answer:

For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.

Step-by-step explanation:

The problem states that the monthly cost of a celular plan is modeled by the following function:

$C(x) = 4 + 0.10(x-70)$

In which C(x) is the monthly cost and x is the number of calling minutes.

How many calling minutes are needed for a monthly cost of at least $7?

This can be solved by the following inequality:

$C(x) \geq 7$

$4 + 0.10(x - 70) \geq 7$

$4 + 0.10x - 7 \geq 7$

$0.10x \geq 10$

$x \geq \frac{10}{0.1}$

$x \geq 100$

For a monthly cost of at least $7, you need to have at least 100 calling minutes.

How many calling minutes are needed for a monthly cost of at most 8:

$C(x) \leq 8$

$4 + 0.10(x - 70) \leq 8$

$4 + 0.10x - 7 \leq 8$

$0.10x \leq 11$

$x \leq \frac{11}{0.1}$

$x \leq 110$

For a monthly cost of at most $8, you need to have at most 110 calling minutes.

For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.

8 0

3 years ago

Please help a girl outtttt

jeyben [28]

Answer:

bottom of graph will move from (0,0) to point (1,3) after transformation

Step-by-step explanation:

given

original : f(x) = $x^{2}$

transformed; g(x) = $(x-1)^{2}$ + 3

look at this way g(x) = $(x-h)^{2}$ + k

if (x-h), h>0, move h units to the right

if k>0, move k units up

the bottom of the graph will be at point (1,3)

3 0

3 years ago