All categories

3 years ago

HELP ME PLEASE!!!!

Mathematics

2 answers:

a_sh-v [17]3 years ago

8 0

For a relation to be a function, it must only have one y-value per x-value.

The top-left answer has two y-values on x=0.
The top-right answer has two y-values on x=-1.
The bottom-left answer has two y-values on x=2.

The answer is the bottom-right graph.

AveGali [126]3 years ago

7 0

Bottom right corner is the correct answer choice

You might be interested in

A recipe uses 8 3/4 cups of milk for every 3 1/2 cups of oatmeal.

Dennis_Churaev [7]

1 1/2 cups of milk for every cup of oatmeal.

8 3/4=8.75 3 1/2=3.50 8.75÷3.50→875÷350=2 175/350

175=1/2 of 350

4 0

3 years ago

11/15-3/5 in fraction

Ket [755]

Answer:

l.c.m of 15 and 5 is 15

next step is to equalize the denominator with the l.c.m value

11/15 is the same (same denominator)

3/5 will change to 9/15 (multiply numerator and denominator by 3 to equalize)

Lastly, you just need to subtract

11/15 - 9/15

2/15

8 0

3 years ago

Find the total surface area of this cuboid, 5 cm 4 cm 6 cm

Tatiana [17]

Answer:

148

Step-by-step explanation:

Sa=2(lw+lh+wh)

Sa=2(4*5+4*6+5*6)

148=2(4*5+4*6+5*6)

3 0

3 years ago

Read 2 more answers

The line represented by the equation 2y - 3x = 4 has a slope of.....

S_A_V [24]

2y - 3x = 4
2y = 3x + 4
y = 3/2*x + 2

y = kx + b ==>

k = 3/2 = 1,5

7 0

4 years ago

A village fete has a children’s running race each year, run in heats of up to ten children. For each heat the first three contes

bekas [8.4K]

Answer: 1) 1/3,654 2) 3/406 3) 72,684,900,288,000 4) 120

Step-by-step explanation:

1) First and Second and Third

$\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}\times \dfrac{1\ remaining\ prize}{27\ remaining\ people}=\dfrac{6}{21,924}\\\\\\.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =\large\boxed{\dfrac{1}{3,654}}$

2) First and Second

$\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}=\dfrac{6}{812}=\large\boxed{\dfrac{13}{406}}$

$3)\quad \dfrac{29!}{(29-10)!}=\large\boxed{72,684,900,288,000}$

$4)\quad _{10}C_3=\dfrac{10!}{3!(10-3)!}=\large\boxed{120}$

8 0

3 years ago