-18(1) = -18,right?
4(3) = 12, right?
-18 - 12 = -30
From the looks of it, we had to "Subtract" the outcome of each problem that we went over
You're correct: -30 (option A.) would be your answer
Answer:
The distance between B and lighthouse is 3.8688 km
Step-by-step explanation:
Given:
The angle made from ship to lighthouse is 36.5 degrees
and that of point B is 73 degrees.
To Find:
Distance Between Point B and Lighthouse
Solution:
<em>Consider a triangle LAB(Refer the attachment )</em>
And Point C is on the line AB as A i.e. ship is sailing to B
So C is at 5 km from A.
Now In triangle LAC,
Using Trigonometry Functions as ,
tan(36.5)=LC/AC
LC=tan(36.5)*AC
=0.7399*5
=3.6998 km
Now In triangle LBC,
As,
Sin(73)=LC/LB
LB=LC/(Sin(73))
=3.6998/0.9563
=3.8688 km
LB=3.8688 km
Answer:
Parallel
Step-by-step explanation:
y=5x+10
y=5x-15
Area=base*height
A=(x+6)(2x+2)
A=2(x+1)(x+6) technically this is the simplified form, but they may want the expanded form...
A=2x^2+14x+12
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.
