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3 years ago

5

Line p has an equation of y=4/9x-3. Perpendicular to line p is line q, which passes through the point (3,-8). What is the equati

on of line q

1 answer:

Mumz [18]3 years ago

3 0

If you look on jiskha you will find the answer you will be looking for just type it in B

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Can you help with this. Thanks!

Ierofanga [76]

92 males buy tickets so
200-92=108
so 108 females buy tickets

30 males buy business and a total of 44 people (female and male) but business so 44-30=14
14 females buy business

108-14(business ticket) = 94 females left.
62 of those 94 females buy economy tickets
so therefore 94-62=32. 32 females left over

14 females = business
62 females = economy
14 + 62 = 76
108 - 76 = 32
32 remaining females buy premium

it says a total of 60 people buy premium so 60 - 32 = 28

so 28 males buy premium

hope this helped

3 0

3 years ago

John Gray bought a basic car for $32,750.00, with options that cost $375.00. There's a 6% sales tax in his state and a combined

alukav5142 [94]

$32,750.00 + $375.00 = $33,125.00
6% of that is $1,987.50 (33,125 x 0.06)
$1,987,50 + $33,125 + $50=
$35,162.50

8 0

3 years ago

Read 2 more answers

The Kendal Company manufactures automobile engine parts. Running the manufacturing operations requires 12 watts (units of electr

Monica [59]

24 because when you plug in 0 for both x and y:

0=12(0)+24

12*0=0, leaving you with 24.

4 0

2 years ago

Unit 3 homework 6 Gina Wilson

Sonja [21]

Answer:

5) The equation of the straight line is 2 x - y + 1 =0

6) The equation of the straight line is x + y -5 =0

7) The equation of the straight line is 5 x + 6 y - 24 =0

8) The equation of the straight line is x - 4 y -4 =0

9) The equation of the parallel line is 3x + y -19 =0

Step-by-step explanation:

5)

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

$m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Given points are (1,3) , ( -3,-5)

$m = \frac{-5-3 }{-3-1 } = \frac{-8}{-4} = 2$

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

$y - 3 = 2 ( x - 1 )$

y = 2x - 2 +3

2 x - y + 1 =0

The equation of the straight line is 2 x - y + 1 =0

6)

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

$m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Given points are (1,4) , ( 6,-1)

$m = \frac{-1-(4) }{6-1 } = \frac{-5}{5} = -1$

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

$y - 1 = -1 ( x - 4 )$

y - 1 = - x +4

The equation of the straight line is x + y -5 =0

7)

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

$m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Given points are (-12 , 14) , ( 6,-1)

$m = \frac{-1-(14) }{6+12 } = \frac{-15}{18} = \frac{-5}{6}$

$y - 14 = \frac{-5}{6} ( x - (-12) )$

6( y - 14 ) = - 5 ( x +12 )

6 y - 84 = - 5x -60

5 x + 6 y -84 + 60 =0

5 x + 6 y - 24 =0

8)

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

$m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Given points are (-4 , -2) , ( 4 , 0)

$m = \frac{0+2}{4 +4} = \frac{2}{8} = \frac{1}{4}$

$y - (-2) =\frac{1}{4} ( x - (-4) )$

4 ( y + 2) = x + 4

x - 4 y -4 =0

9)

The equation of the line y = 3x + 6 is parallel to the line

3x + y + k =0 is passes through the point ( 4,7 )

⇒ 3x + y + k =0

⇒ 12 + 7 + k =0

⇒ k = -19

The equation of the parallel line is 3x + y -19 =0

4 0

2 years ago

Find the value of f(9) and g(-9) if f(×) = -7×-9 and g(×) = 6×3-23×

Vlada [557]

Answer:

f(9)= -72

g(-9)= 225

Step-by-step explanation:

First put the original equation

f(x)=-7x-9

Next plug in the number f(9)

f(9)=-7(9)-9

After multiply -7(9) which gets you -63

f(9)=-63-9

Then subtract -63-9 which gets you -72

f(9)=-72

-----------------------------------------------------------

First put the original

g(x)=6x3-23x

Next plug in the number g(-9)

g(-9)=6x3-23(-9)

After multiply 6x3 which gives you 18 then multiply -23(-9)

g(-9)=18+207

then add 18+207 which gives you 225

g(-9)=225

8 0

3 years ago