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sergey [27]
3 years ago
11

Which relations are functions?function or not a function?

Mathematics
1 answer:
Bad White [126]3 years ago
6 0

Hi there,

The first one is a function this is because it has a straight line and you are able to figure out the slope.

The second one is not a function because it is not a straight line and with that you are unable to determine the slope.

The third one is a function. It has a straight line.

Lastly, the fourth one is not a function. It doesn't have a straight line

Hope these are correct :)

Have a great day

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What is the answer to the problem 2/3x-40=5/6(x+36)
miss Akunina [59]

2/3x-40=5/6(x+36)

multiply each side by 6 to get rid of the fraction

6* 2/3x- 6*40=6*5/6(x+36)

4x -240 = 5(x+36)

distribute

4x-240 = 5x +180

subtract 4x from each side

-240 = x +180

subtract 180 from each side

-420 =x

5 0
3 years ago
After her raise, Shannon earned 3/4 more than she previously made. If Shannon's weekly wages totaled $2100 after her raise, what
topjm [15]
So $2100 = 1+3/4 of her original raise.

Remember that 1 = 4/4, so in total it is 7/4 of her original pay (4/4+3/4). Let's find out what 1/4 is equal to.

If $2100=7/4 then 1/4=$300. Therefore 4/4 (or 1, her original pay) is 4 times $300, which is $1200.
8 0
3 years ago
Jacob designs a building with a square base. He starts by building a model for the building, which has a base with an
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Answer:

Step-by-step explanation:

base of model is 6 ft by 6 ft

base of building is 78 ft by 78 ft

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8 0
3 years ago
20. Which set of parametric equations over the interval 0 ≤ t ≤ 1 defines a line segment with initial point (–5, 3) and terminal
lesya [120]

Given:

Initial point = (–5, 3)

Terminal point = (1, –6)

To find:

The set of parametric equations over the interval 0 ≤ t ≤ 1.

Solution:

The interval is 0 ≤ t ≤ 1, initial point is (–5, 3) and terminal point is (1, –6). It means,

x(0)=-5,y(0)=3

x(1)=1,y(1)=-6

Put t=1 in each parametric equation.

In option A,

x(1)=-5+1=-4\neq 1

In option B,

x(1)=-5+3(1)=-5+3=-2\neq 1

In option D,

x(1)=-5+8(1)=-5+8=3\neq 1

Therefore, options A, B and D are incorrect.

In option C,

x(1)=-5+6(1)=-5+6=1

y(1)=3-9(1)=3-9=-6

Put t=0 in x(t) and y(t).

x(0)=-5+6(0)=-5

y(0)=3-9(0)=3

Since, only in option C x(0)=-5,y(0)=3,x(1)=1,y(1)=-6, therefore, the correct option is C.

7 0
3 years ago
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viva [34]
It is <span> n = -14 i believe</span>
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