Answer:
The common difference is -5/4
T(n) = T(0) - 5n/4,
where T(0) can be any number. d = -5/4
Assuming T(0) = 0, then first term
T(1) = 0 -5/4 = -5/4
Step-by-step explanation:
T(n) = T(0) + n*d
Let
S1 = T(x) + T(x+1) + T(x+2) + T(x+3) = 4*T(0) + (x + x+1 + x+2 + x+3)d = 240
S2 = T(x+4) + T(x+5) + T(x+6) + T(x+7) = 4*T(0) + (x+5 + x+6 + x+7 + x+8)d = 220
S2 - S1
= 4*T(0) + (x+5 + x+6 + x+7 + x+8)d - (4*T(0) + (x+1 + x+2 + x+3 + x+4)d)
= (5+6+7+8 - 1 -2-3-4)d
= 4(4)d
= 16d
Since S2=220, S1 = 240
220-240 = 16d
d = -20/16 = -5/4
Since T(0) has not been defined, it could be any number.
Answer:9.9
Step-by-step explanation:
Add up the scores and dividing the total by the number of scores.
24 pens divided by 2 boxes equals 12 pens per box.
Answer:
-x^2+15y-15
Answer if it is a ^2
Step-by-step explanation:
2x^2 + 6y - 3 -(3x^2 - 9y + 12)=
2x^2+6y-3-3x^2+9y-12=
-x^2+15y-15
is it a ^2 or a question mark
Answer:
I) |xz| ≈ 28.6 km
II) |yz| ≈ 34.8 km
Step-by-step explanation:
Let's assume that the position of ship due south of x is z (aà pictor representation of the question is attached)
|xy| = 20 km, |xz| = ?, |yz| = ?, θ(y) = 55°
Using Trigonometric ratio - SOHCAHTOA
I) Tan θ = |xz| ÷ |xy| ⇒ Tan 55° = |xz| ÷ 20
|xz| = 20 * Tan 55 = 20 * 1.428
|xz| = 28.56 km
|xz| ≈ <u>28.6 km</u>
<u />
II) Cos θ = |xy| ÷ |yz| ⇒ Cos 55° = 20 ÷ |yz|
|yz| * Cos 55° = 20 ⇒ |yz| = 20 ÷ Cos 55°
|yz| = 20 ÷ 0.574 = 34.84 km
|yz| ≈ <u>34.8 km</u>
