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3 years ago

Which table represents a linear function? PLEASE HELP ME

Mathematics

1 answer:

-Dominant- [34]3 years ago

6 0

Answer:

Bottom left

Step-by-step explanation:

If you graph out the coordinates it goes in a perfect line, thus being linear.

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To win a particular lottery game, it is necessary to match numbers on 5 balls that are randomly picked from 65 balls numbered 1-

emmainna [20.7K]

Answer: $5.34*10^{-5}$

Step-by-step explanation:

given data:

number of balls = 65.

special black ball that is picked from 10 other balls = numbered 0-9.

Solution:

probability that a single lottery ticket will match 2 balls and the special black ball.

$= 65C2*9C1\\\\= 18720\\\\=\frac{1}{18720} \\\\= 5.34*10^{-5}$

5 0

3 years ago

Find the coordinates of the missing endpoint if M is the midpoint of DF F(2,9) M(-1,6)

Hunter-Best [27]

Answer:

(-4,3)

Step-by-step explanation:

6 0

3 years ago

Findes the slope from two points point 1=(4,5) point 2=(6,9)

hoa [83]

Answer:

Step-by-step explanation:

The formula for findining slope it y2-y1/x2-x1

so 9-5=4/ 6-4=2

4/2 is 2

the slope is 2

6 0

4 years ago

Sandy is helping her uncle at this coffee shop. She observes that, when ordering coffee, 13 of customers select chocolate flavor

Sophie [7]

19/50 people will probably order chocolate

3 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago