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evablogger [386]

3 years ago

5

Drag and drop each pair of lines into the correct category to indicate whether the pair of lines are parallel, perpendicular, or

neither.

1 answer:

zhenek [66]3 years ago

6 0

parallel:

y=5x-1 ; 10x-2y=12

perpendicular:

3x-5y=-1 ; 3y=-5x+2

6y=4x+1 ; 12x+8y=12

neither:

1/7x+y=7 ; y=-7x

x-6y=12 ; x+4

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I really need help who can help me?

Gelneren [198K]

Yes
Y varies directly with x, it is a direct relationship between the two.

3 0

3 years ago

Which equation can be used to find the number of fish in the pond on June 1?

olga_2 [115]

Answer: The required equation is $f-52=63$ and the number of fishes in the pond on June 1 = 115

Step-by-step explanation:

Given: The number of fishes in pond on July 1 = 63

Also, it is said that the number of fishes in pond on July 1 is 52 fewer(lesser) than the number of fishes in pond on June 1.

Let f be the number of fishes in pounds on June 1.

then the required equation will be

$f-52=63$

Now, add 52 on both the sides of the equation, we get

$f=63+52\\\Rightarrow\ f=115$

3 0

4 years ago

M is directly proportional to r2 <br> When r=2, m=14<br> Work out the value of m when r=12

Genrish500 [490]

we have M is durectly porpotional to r^2

so M=(k)r^2

and when r=2, m=14

so 14=(k)(2)^2

k=14/4 =7/2

so when r=12

m= (7/2)(12)^2 =(7/2)(144) = 504

8 0

3 years ago

Show that the following functions are probability density functions for some value of k and determine k. Then, determine the mea

lord [1]

Answer:

a) 17.5

b) 15.6

c) 13.3

d) 21.51

Step-by-step explanation:

The given function is equal to:

f(x)=kx^2

where

$\int\limits^y_0 {kx^{2} } \, =1$

where y=23

Clearing k=0.00025

a) $Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.00025x^{2} } \, dx =17.5$

b) $Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2}f(x) } \, dx-17.5^{2} =\int\limits^y_0 {x^{2} *0.00025x^{2} } \, dx -17.5^{2} =321.82-306.25=15.6$

c) The function is equal to:

f(x)=k(1+2x)

$\int\limits^y_0 {k(1+2x)} \, =1$

where y=20

k=0.0024

$Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.0024(1+2x)} \, dx =13.3$

d) $Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2} f(x)} \, dx -13.3^{2}=\int\limits^y_0 {x^{2} *0.0024(1+2x)} \, dx-13.3^{2} =198.4-176.89=21.51$

8 0

4 years ago

Find the coordinates and midpoint of H(0,0) and X(8,4)

sergejj [24]

Divid the differince of the x's by 2 and the diffence of the y's also by 2
8/2,4/2 is (4,2)
M=(4,2)

hope this helps
:)

7 0

3 years ago