Answer:
Acute angle.
Right angle.
Obtuse angle.
Straight angle.
Reflex angle
Step-by-step explanation:
(a) The "average value" of a function over an interval [a,b] is defined to be
(1/(b-a)) times the integral of f from the limits x= a to x = b.
Now S = 200(5 - 9/(2+t))
The average value of S during the first year (from t = 0 months to t = 12 months) is then:
(1/12) times the integral of 200(5 - 9/(2+t)) from t = 0 to t = 12
or 200/12 times the integral of (5 - 9/(2+t)) from t= 0 to t = 12
This equals 200/12 * (5t -9ln(2+t))
Evaluating this with the limits t= 0 to t = 12 gives:
708.113 units., which is the average value of S(t) during the first year.
(b). We need to find S'(t), and then equate this with the average value.
Now S'(t) = 1800/(t+2)^2
So you're left with solving 1800/(t+2)^2 = 708.113
<span>I'll leave that to you</span>
1. The growth rate equation has a general form of:
y = A (r)^t
The function is growth when r≥1, and it is a decay when
r<1. Therefore:
y=200(0.5)^2t -->
Decay
y=1/2(2.5)^t/6 -->
Growth
y=(0.65)^t/4 -->
Decay
2. We rewrite the given equation (1/3)^d−5 = 81
Take the log of both sides:
(d – 5) log(1/3) = log 81
d – 5 = log 81 / log(1/3)
d – 5 = - 4
Multiply both sides by negative 1:
- d + 5 = 4
So the answer is D
Answer:
5 candies
Step-by-step explanation:
Total of candy Sarah has = 25candies
If Sarah gave 2/5 of the candies to her sister, the total amount she gave her sister will be 2/5 of 25
2/5 of 25 = 10
The remaining candy left will be 25-10 which is 15.
If she gave 2/3 of amount left to her friend, the amount she gave her friend will be 2/3×15 = 10candies.
The amount of candy Sarah have left will be 15- 10 which is 5candies.
4.a) 2x=7*3
2x=21
x=21/2 OR
x=10.5
4.b)5x=25*6
5x=150
x=150/5
x=30
4.c)4x=5*14
4x=70
x= 70/4
x=17.5
4.d)4x=9*3
4x=27
X=27/4
X=6.75
4.e)x=0.05*3
X=0.15
4.f)x=0.25*7
X=1.75
4.g)x=0.4*1.5
X=0.6
4.h)x=0.7*2.2
X=1.54
5a)x/7=7-3
X/7=4
X=4*7
X=28
B)2x/5=3+8
2x/5=11
2x=11*5
2x=55
X=55/2
X=27.5
C)2x/3=74-26
2x/3=48
2x=48*3
2x=144
X=72
(I can’t finish all but it is the same method )
