Answer:
James will end up with his original t cars and half of (t+13) cars, so will have ...
... t + (t+13/2) = (3t +13)/2 . . . . cars James has after Paul's gift
Step-by-step explanation:
If we divide 56 by 4 we get 14. Why by4? So that one number is 3 times the other So he had 14 balls that went into the green bags and 42 (56-14) that went into the red bags. We could just answer the question and say 14 but I think they want to know how many in each green bag.
14 and 42 don't work because they are not the same number of balls. What number is a common factor? 7 is,
We could have 2 green bags and split the 14 balls into 2 groups of 7 and with the remaining 42 - put them into 6 red bags of 7 each.
And so the answer to your question is:
7 ball in each bag = 2 bags are green, and 6 bags are red
14 balls + 42 balls = 56 bouncy balls
Answer:
y = 6x - 11
Step-by-step explanation:
An arithmetic sequence is a linear function. This means it steadily increases or decreases through addition/subtraction by a constant amount. Here the sequence grows by adding 6 each time.
-11 + 6 = -5
-5 + 6 = 1
1 + 6 = 7
etc....
Since this is linear, its equation has the form y = mx + b where m is the slope and b is the y-intercept. The slope is m = 6 and b = -11.
y = 6x - 11
Answer:
The diagram for the question is missing, but I found an appropriate diagram fo the question:
Proof:
since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle
∠BCO = 45°
∠BOC = 45°
∠PCO = 45°
∠POC = 45°
∠DOP = 22.5°
∠PDO = 67.5°
∠ADO = 22.5°
∠AOD = 67.5°
Step-by-step explanation:
Given:
AB = CD = 297 mm
AD = BC = 210 mm
BCPO is a square
∴ BC = OP = CP = OB = 210mm
Solving for OC
OCB is a right anlgled triangle
using Pythagoras theorem
(Hypotenuse)² = Sum of square of the other two sides
(OC)² = (OB)² + (BC)²
(OC)² = 210² + 210²
(OC)² = 44100 + 44100
OC = √(88200
OC = 296.98 = 297
OC = 297mm
An isosceless tringle is a triangle that has two equal sides
Therefore for △OCD
CD = OC = 297mm; Hence, △OCD is an isosceless triangle.
The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles
Since BC = OB = 210mm
∠BCO = ∠BOC
since sum of angles in a triangle = 180°
∠BCO + ∠BOC + 90 = 180
(∠BCO + ∠BOC) = 180 - 90
(∠BCO + ∠BOC) = 90°
since ∠BCO = ∠BOC
∴ ∠BCO = ∠BOC = 90/2 = 45
∴ ∠BCO = 45°
∠BOC = 45°
∠PCO = 45°
∠POC = 45°
For ΔOPD

Note that DP = 297 - 210 = 87mm
∠PDO + ∠DOP + 90 = 180
∠PDO + 22.5 + 90 = 180
∠PDO = 180 - 90 - 22.5
∠PDO = 67.5°
∠ADO = 22.5° (alternate to ∠DOP)
∠AOD = 67.5° (Alternate to ∠PDO)