-t=9t(t-10) Im stuck

Consider the line 5x-8y=8 what is the slope of a line perpendicular to this line what is the slope of a line parallel to this li

Sophie [7]

$\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\\text{Convert the equation to the slope-intercept form}\ y=mx+b.\\\\5x-8y=8\qquad\text{subtract}\ 5x\ \text{from both sides}\\\\-8y=-5x+8\qquad\text{divide both sides by (-8)}\\\\y=\dfrac{5}{8}x-1\to m_1=\dfrac{5}{8}\\\\perpendicular:\ m_2=-\dfrac{1}{\frac{5}{8}}=-\dfrac{8}{5}\\\\parallel:\ m_2=\dfrac{5}{8}$

8 0

3 years ago

Philip tiene 8 bolas rojas, 3 bolas verdes, 6 bolas amarillas, 13 bolas negras y 15 bolas azules en su bolsa. ¿cuál es la media,

Zielflug [23.3K]

Answer:

Media: 9
Mediana: 8
Moda: 8, 3, 6, 13, 15
Rango: 12

Step-by-step explanation:
Ojala que te ayude ^^

3 0

2 years ago

A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orang

laiz [17]

Answer-

a. Probability that three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

Solution-

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

$\binom{19}{3} = \frac{19!}{3!16!} = 969$

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

And this process can be done in,

$\binom{33}{6} = \frac{33!}{6!27!} =1107568$

So total number of selection = (969)×(1107568) = 1073233392

Drawing 9 candies out of 52 candies,

$\binom{52}{9} = \frac{52!}{9!43!} = 3679075400$

∴P(3 white candies) = $\frac{1073233392}{3679075400} =0.29$

Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,

$\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}$

$=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})$

$=(969)(45)(7)(5)(15)=22892625$

Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,

$\binom{52}{9}=\frac{52!}{9!43!} =3679075400$

∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =

$\frac{22892625}{3679075400} = 0.006$

6 0

3 years ago

ILL BRAINLIEST YOU PLEASE HELP ME

Aleksandr-060686 [28]

Answer:

x = 71/5 (as an improper fraction)

x = 14 1/5 (as a proper fraction

x = 14.2 (as a decimal)

Step-by-step explanation:

It's a rectangle so the diagonals are equal

7x - 9 = 2x + 62

Subtract 2x from both sides

5x - 9 = 62

Add 9 to both sides

5x = 71

Divide both sides by 5

x = 71/5 (as an improper fraction)

x = 14 1/5 (as a proper fraction

x = 14.2 (as a decimal)

8 0

2 years ago

Read 2 more answers

Which of the following vectors are orthogonal to (1,5)? Check all that apply

Andre45 [30]

We know that
If the scalar product of two vectors is zero, both vectors are orthogonal

A. (-2,5)
(-2,5)*(1,5)-------> -2*1+5*5=23-----------> are not orthogonal

B. (10,-2)
(10,-2)*(1,5)-------> 10*1-2*5=0-----------> are orthogonal

C. (-1,-5)
(-1,-5)*(1,5)-------> -1*1-5*5=-26-----------> are not orthogonal

D. (-5,1)
(-5,1)*(1,5)-------> -5*1+1*5=0-----------> are orthogonal

the answer is
B. (10,-2) and D. (-5,1) are orthogonal to (1,5)

7 0

3 years ago