1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Valentin [98]
3 years ago
11

At Thursday’s basketball game, there were more than twice as many adult spectators as children. If x represents the number of ad

ults and y represents the number of children, which inequality can be used to find the possible combinations of adults and children at the game?
Mathematics
2 answers:
elixir [45]3 years ago
7 0
The inequality equation should be <span>x > 2y.</span>
larisa [96]3 years ago
7 0

Answer:

x > 2y

Step-by-step explanation:

It is given in the question that there were more than twice as many adult spectators as children.

If x represents the number of adults and y represents the number of children then inequality representing the relation will be

x > 2y

So inequality given above can be used to find the possible combinations of adults and children at the game.

You might be interested in
Rewrite 3x+4y=12 in slope-intercept form. Then graph the equation.
frez [133]

Answer:

3x+4y=12

Step-by-step explanation:

7 0
2 years ago
CALLING ALL BRAINLY USER HELP ME AND I'LL MARK U (BRAINLLEST)!!!
Natalka [10]

Answer:

it would be b for 1 since 7 would be negative and the 3 would be positive

7 0
3 years ago
The length of a lap is 20 meters. if lisa wants to swim 800 meters this week how many laps must she swim
marishachu [46]
40 laps/////////////
5 0
3 years ago
Given y=x2 + 9x + 20 find the zeroes.
Artyom0805 [142]

Answer:

x=-5   x=-4

Step-by-step explanation:

y=x^2 + 9x + 20

We need to factor the equation

What 2 numbers multiply to 20 and add to 9

4*5 = 20

4+5 =9

y=(x+5) (x+4)

To find the zero's we set the equation equal to zero

0= (x+5) (x+4)

Using the zero product property

x+5 = 0   x+4=0

x=-5   x=-4

3 0
3 years ago
Tentukan persamaan garis yang di wakili ruas garis AB,BC,CD dan AD! <br>tolong kak
Ann [662]

Answers:

1. AB: y = - 2x - 6

2. BC: y = x - 3

3. CD: x = 3

4. AD: y = (3/7) x +26/7


Solution:

y-y1=m(x-x1)

m=(y2-y1)/(x2-x1)


1. AB

A=(-4,2)=(xa, ya)→xa=-4, ya=2

B=(-1,-4)=(xb, yb)→xb=-1, yb=-4

mab=(yb-ya)/(xb-xa)

mab=(-4-2)/(-1-(-4))

mab=(-6)/(-1+4)

mab=(-6)/(3)

mab=-2

y-ya=mab (x-xa)

y-2=-2(x-(-4))

y-2=-2(x+4)

y-2=-2x-8

y-2+2=-2x-8+2

y=-2x-6


2. BC

B=(-1,-4)=(xb, yb)→xb=-1, yb=-4

C=(3,0)=(xc, yc)→xc=3, yc=0

mbc=(yc-yb)/(xc-xb)

mbc=(0-(-4))/(3-(-1))

mbc=(0+4)/(3+1)

mbc=(4)/(4)

mbc=1

y-yc=mbc (x-xc)

y-0=1 (x-3)

y=x-3


3. CD

C=(3,0)=(xc, yc)→xc=3, yc=0

D=(3,5)=(xd, yd)→xd=3, yd=5

mcd=(yd-yc)/(xd-xc)

mcd=(-4-0)/(3-3)

mcd=(-4)/(0)

mcd= Infinite (Vertical line)

Equation vertical line: x=xc=xd

x=3


4. AD

A=(-4,2)=(xa, ya)→xa=-4, ya=2

D=(3,5)=(xd, yd)→xd=3, yd=5

mad=(yd-ya)/(xd-xa)

mad=(5-2)/(3-(-4))

mad=(3)/(3+4)

mad=3/7

y-ya=mad (x-xa)

y-2=(3/7)(x-(-4))

y-2=(3/7)(x+4)

y-2=(3/7)x+12/7

y-2+2=(3/7)x+12/7+2

y=(3/7)x+(12+7(2))/7

y=(3/7)x+(12+14)/7

y=(3/7)x+26/7


8 0
3 years ago
Other questions:
  • Complete the inequality statement.<br><br> 0.83 ___ 8.3%
    15·2 answers
  • Write a polynomial
    12·1 answer
  • For any two nonzero integers, the product and quotient have the same sign. True or False
    11·2 answers
  • Please answer this now with correct answer
    12·1 answer
  • Solve for y: (1/2)y + 1/2 =<br> 3 1/2
    14·1 answer
  • Fill in the missing verb forms
    12·1 answer
  • What is the volume of the rectangular prism? A prism has a length of 8 inches, width of 2 inches, and height of 12 and one-half
    7·2 answers
  • Which equation is a function of x?
    14·2 answers
  • 4) Daily tickets to Six Flags cost $50 each.<br> Variable #1:<br> What is the variable
    11·1 answer
  • Draw the line of reflection that reflects AABC onto AA'B'C'. please help!!
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!