Ask question

Ask question

All categories

3 years ago

7

Find the range of the relation {(–1, 4), (2, 5), (3, 5)}. Then determine whether the relation is a function.

1 answer:

sergiy2304 [10]3 years ago

3 0

The range is the y values of the function. So it would be (4,5). In order for the relation to be a function, it has to pass the vertical line test. That means that one singular x value cannot equal two different y values. It makes no sense if you really think about it. You can’t plug two different x values into the same function and get the same y value. There’s no repeat in x values, so the relation is a function

You might be interested in

Which of the following is most likely the next step in the series ?

GalinKa [24]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Which of the following is most likely the next step in the series ?

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

The next series is most probably going to be <u>option </u><u>D.</u><u> </u>
We can see from the series that, figure 1 has 4 points marked on the circle from which they formed a rectangle.
In the next figure, there are 3 points marked which together when joined forms a triangle.
Then the next circle has 2 points marked on it which makes a straight line.
So, obviously the next figure/circle in the series should have only 1 point marked on it. This won't form anything as you can't join just 1 point with something.

7 0

2 years ago

Read 2 more answers

The quotient of n and 6

Shalnov [3]

Answer:

The quotient of n and 6 is n/6, as quotient means divide.

for example, if it was "the quotient of 24 and 6", it would be 24/6, which is 4.

5 0

2 years ago

How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

7 0

2 years ago

Write an equation for a line passing through (-4, 3) and (0, 6)

Elena L [17]

Answer:

y = 3/4x + 6

Step-by-step explanation:

6-3 3

------- = ------

0--4 4

y = mx + b

take coordinates for y and x out of (-4, 3) or (0,6)

6= 3/4 * 0 + b

6 = 0 + b

b = 6

7 0

2 years ago

This is very easy!!

Olenka [21]

Answer:

product

Step-by-step explanation:

hope it helps

5 0

2 years ago

Read 2 more answers