X = 0 ; y = 10 ; 10 = a(0) + b(0) + c
c = 10
x=2 ; y = 15 ; 15 = a(4) + b(2) + 10 ; 5 = 4a+2b
x=4 ; y=18 ; 18 = a(16) + b(4) + 10 ; 8 = 16a + 4b
2(5) = (4a+2b)2
-
<u> 8 = 16a + 4b
</u> 2 = -8a
a = -0.25
b = 2
y = (1/4)x^2 + 2x + 10 ; 4y = x^2 + 8x + 40
Answer:
< CFE = 40°
Step-by-step explanation:
To better understand the solution, see attachment for the diagram.
Given:
BC parallel to DE
Measure of Arc BD = 58°
Measure of Arc DE = 142°
First step: Draw a diameter that passes through the centre of the circle and name it. In this case, the diameter is line ST.
The line ST divides the arc BD and arc DE into half.
That is:
Arc SC = 1/2(arc BC) =1/2(58)
Arc SC = 29°
Arc TE = 1/2(arc DE) =1/2(142)
Arc TE = 71°
Arc SC + Arc CE + Arc TE = 180° (Sum of angles in a semicircle
29° + Arc CE + 71° = 180°
Arc CE + 100° = 180°
Arc CE = 180-100
Arc CE = 80°
Inscribed angle = 1/2(intercepted angle)
<CFE = 1/2(Arc CE )
<CFE = 1/2(80)
< CFE = 40°
Let us make a list of all the details we have
We are given
The cost of each solid chocolate truffle = s
The cost of each cream centre chocolate truffle = c
The cos to each chocolate truffle with nuts = n
The first type of sweet box that contains 5 each of the three types of chocolate truffle costs $41.25
That is 5s+5c+5n = 41.25 (cost of each type of truffle multiplied by their respective costs and all added together)
The second type of sweet box that contains 10 solid chocolate trufles, 5 cream centre truffles and 10 chocolate truffles with nuts cost $68.75
That is 10s+5c+10n = $68.75
The third type of sweet box that contains 24 truffles evenly divided that is 12 each of solid chocolate truffle and chocolate truffle with nuts cost $66.00
That is 12s+12n=$66.00
Hence option C is the right set of equations that will help us solve the values of each chocolate truffle.
