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3 years ago

Determine the domain of the relation. {(3.0), (2,8), (17,10)}

Mathematics

1 answer:

Trava [24]3 years ago

3 0

Answer:

work is pictured and shown

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Which statement is used to prove that quadrilateral abcd is a parallelogram

gizmo_the_mogwai [7]

Answer:

Opposite sides must be congruent.

Step-by-step explanation:

Given is a picture of a quadrilateal ABCD

We have to prove that this is a parallelogram.

Recall the properties of parallelogram as

i) diagonals bisect each other

ii) opposite sides parallel

iii) opposite sides congruent

or iv) one pair of opposite sides equal and parallel

We have hence to get opposite sides congruent

So last option is right

5 0

3 years ago

Can someone please help me? I can't find the h(x) for 0. I also need help with the last question. Please I need it very soon.

Solnce55 [7]

9514 1404 393

Answer:

see attachments for a table and graph

Step-by-step explanation:

I find it convenient to use a graphing calculator or spreadsheet to do repetitive computations reliably.

Your specific question asks for h(0):

h(0) = -|0 +2| +1 = -2 +1

h(0) = -1

You may want to check your other table values.

8 0

3 years ago

36 inches/second = yards/minute<br><br> How many yards?

gulaghasi [49]

Answer:

1 yard

Step-by-step explanation:

3 0

3 years ago

4 people went to the movies. Each person spent $9.50 on a ticket and had the same amount of money leftover. They originally had

solong [7]

$2.50 was left for each person.

4(9.50) + x = 48

38 + x = 48

subtract 38 on both sides

x = 10

10$ was left in total by all 4 people

10/4 = 2.50

$2.50 was left over for each person

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

<u>Step-by-step explanation:</u>

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