Answer:
He must win 11 more matches to qualify for the bonus.
Step-by-step explanation:
24/36 * 100 = 67% (to the nearest %) - This is the current percentage of what the Tennis player has won.
If we add 14 more matches on to the 36 the player has already played, we know that the Tennis player plays 50 matches in total.
Let's say the Tennis player was playing 100 matches, they would need to win 70 or more to qualify for the bonus. Because the player is playing half this amount of matches, we half the amount of games they have to win...
35/50 games or more must be won to qualify for the bonus. The Tennis player has already won 24 matches, so must win 11 more matches to qualify for the bonus.
Hope that helps!
Do you have any more information???
If this were to be graphed, the independent variable would be the price of the ticket for the rides. The dependent variable would be the total cost.
The fair admission is not a variable because it is a constant price for every single person who goes into the fair.
The problem asks to use y to represent the total cost and x to represent the number of ride tickets. In order to fully write out the equation, we have to figure out what the fair admission costs.
43.75 = 1.25(25) + b
*b represents the fair admission
Multiply 1.25 by 25
43.75 = 31.25 + b
Subtract 31.25 to find what b costs.
12.50 = b
The fair admission costs $12.50.
Solution: y = 1.25x + 12.50
Answer:
0.6
Step-by-step explanation:
3+2=5
P (white) = 3/5 = 0.6
(Hopefully this works, if not, on hearty maths you can go back to your assigned tasks page and go back onto the task to get a new question if that makes sense)
<em>Given - a+b+c = 0</em>
<em>To prove that- </em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
<em>Now we know that</em>
<em>when x+y+z = 0,</em>
<em>then x³+y³+z³ = 3xyz</em>
<em>that means</em>
<em> (x³+y³+z³)/xyz = 3 ---- eq 1)</em>
<em>Lets solve for LHS</em>
<em>LHS = a²/bc + b²/ac + c²/ab</em>
<em>we can write it as LHS = a³/abc + b³/abc + c</em><em>³</em><em>/abc</em>
<em>by multiplying missing denominators,</em>
<em>now take common abc from denominator and you'll get,</em>
<em>LHS = (a³+b³+c³)/abc --- eq (2)</em>
<em>Comparing one and two we can say that</em>
<em>(a³+b³+c³)/abc = 3</em>
<em>Hence proved,</em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
