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Mathematics 9 Online
OpenStudy (anonymous):

Medal + Fan Tickets to a basketball game can be ordered online for a set price per ticket plus a $5.50 service fee. The total cost in dollars for ordering 5 tickets is $108.00. Which linear function represents c, the total cost, when x tickets are ordered? c(x) = 5.50 + 20.50x c(x) = 5.50x + 20.50 c(x) = 5.50 + 21.60x c(x) = 5.50x + 21.60

OpenStudy (anonymous):

@Haseeb96

OpenStudy (anonymous):

I got C?

OpenStudy (haseeb96):

how did you take out 21.60 ??

OpenStudy (anonymous):

subtracted

OpenStudy (haseeb96):

can you plz do it here .? if so do quickly

OpenStudy (anonymous):

I only havea few minutes.. could you show me please I'll find out my mistake

OpenStudy (anonymous):

I'm sorry i'm just running low on time

OpenStudy (haseeb96):

are u sure that it will be 21.60 ? if so then D is correct answer

OpenStudy (haseeb96):

no C is correct answer

OpenStudy (anonymous):

im not 100%

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

I got one more, I know how to do this type of problem but not in standard form

OpenStudy (anonymous):

The point-slope form of the equation of the line that passes through (–9, –2) and (1, 3) is y – 3 = (x – 1). What is the slope-intercept form of the equation for this line?

OpenStudy (haseeb96):

m=3+2/1+9 = 5/10 = 1/2 m=1/2 now put any point suppose i choose (1,3) and put it in this formula (y-y1) =m(x-x1) y-3=1/2 (x-1) 2y-6=x-1

OpenStudy (haseeb96):

2y = x-1+6 2y= x+5 y= x/2 +5/2

OpenStudy (haseeb96):

this is your answer

OpenStudy (anonymous):

got it one more same type

OpenStudy (anonymous):

Line CD passes through points C(3, –5) and D(6, 0). What is the equation of line CD in standard form? 5x + 3y = 18 5x – 3y = 30 5x – y = 30 5x + y = 18

OpenStudy (anonymous):

last one

OpenStudy (haseeb96):

m=5/3 (x1,y1) = (6,0) y-0 =5/3 (x-6) 3y = 5x - 30 y= 5x/3 - 10

OpenStudy (anonymous):

Thank you!!!!!

OpenStudy (anonymous):

thank you!

OpenStudy (anonymous):

Best helper i've found on here

OpenStudy (haseeb96):

welcome welcome :) whenever you need help just tag me

OpenStudy (anonymous):

thanks once again, I will. I might come to you tomorrow if that's alright

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