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fomenos
3 years ago
12

Select the sets that are not functions.

Mathematics
1 answer:
DedPeter [7]3 years ago
5 0

Answer:

Sets D, E

Step-by-step explanation:

Sets D and E both have repeated x-values , thus making them a relation not a function.

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A car travels a distance of 500 km in a time of 5 hours. <br> What is the car's average speed?
FrozenT [24]

Answer:

100km/h

Step-by-step explanation:

The formula for speed is:

v  =  x / d  =  500/5  =  100km/h

7 0
3 years ago
Please help! I really need help...
Sholpan [36]

Answer:

3y^{2}+2y-11

Step-by-step explanation:

we have

(-y^{2}-4y-8)-(-4y^{2}-6y+3)

Eliminate the parenthesis

-y^{2}-4y-8+4y^{2}+6y-3

Group terms that contain the same variable

-y^{2}+4y^{2}-4y+6y-8-3

Combine like terms

3y^{2}+2y-11

4 0
4 years ago
60% of the students at a school walk to school. Work out the percentage of the students who do not walk ​
Shkiper50 [21]

Answer:

40%

Step-by-step explanation:

everything is 100% so you have to

100-60=40

40% of people do not walk to school

3 0
2 years ago
Lee's family had a garden in the shape of a right triangle. the total area of the garden is 16 square feet. if they expand the g
lana [24]
Area of right-angled triangle  is given by;

Area, A = 1/2 *b*h, Where b=base, h=height

Therefore,

A1 = 1/2bh = 16 in^2
A2 = 1/2 (2b)(2h) = 2bh

Ratio of increase = A2/A1 = {2bh}/(1/2bh} = 4 (the area is increased 4 times)
The,
A2 = 4*16 = 64 in^2
Therefore,
The area is increased by (64-16) = 48 in^2
5 0
3 years ago
Find Vertical Asymptote(s) for y= (4x^2 +1) / (x^2 -1)
Novay_Z [31]

Answer:

Vertical asymptotes are at x=-1\ and\ x=1.

Step-by-step explanation:

The rational function is given as:

y=\frac{4x^2+1}{x^2-1}

Vertical asymptotes are those values of x for which the function is undefined or the graph moves towards infinity.

For a rational function, the vertical asymptotes can be determined by equating the denominator equal to zero and finding the values of x.

Here, the denominator is x^2-1

Setting the denominator equal to zero, we get

x^2-1=0\\x^2=1\\x=\pm \sqrt{1}\\ x= -1\ and\ x = 1

Therefore, the vertical asymptotes occur at x=-1\ and\ x=1.

6 0
3 years ago
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