Hello
1 ) 3+4sinQ=6 ( Q is theta)
4sinQ=6 -3
sinQ=3/4 conclusion : Q= Arcsin(3/4)=......calculator
2 ) (2/7)cosQ =1/4
cosQ = 7/8 conclusion : Q= Arcsin(7/8)=....calculator
So basically you want to find the total percentage of cakes that weren't sponge cakes first.
You already have the 50% of party cakes. For the 1/5 percent of fruit cakes, you can multiply it by 20/20, to get 20/100, and then just take the 20 to get 20% that were fruit cakes.
Now you can just add the percentages together.
50% + 20% = 70%
So now you know 70% weren't sponge cakes, out of 100%.
So here you can just subtract 70% from 100% to figure out the remaining part of 100%, which must be sponge cakes.
100% - 70% = 30%
So 30% of the cakes were sponge cakes.
Well he replace 1 of 3 then he has 3 and if he replaced a total of all three then he would have 3 I'm not all that sure tho sorry this question is kinda confusing
Problem 1
We replace every copy of x with 4c and simplify like so:
f(x) = 8 - 5x
f(4c) = 8 - 5*4c
f(4c) = 8 - 20c is the answer
This is equivalent to -20c+8.
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Problem 2
Same idea as before, but this time we'll plug in x = 2-k
Each x gets replaced with (2-k)
f(x) = 8 - 5x
f(2-k) = 8 - 5(2-k)
f(2-k) = 8 - 10 + 5k
f(2-k) = -2 + 5k
This is the same as saying 5k - 2.
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Problem 3
The steps are similar to earlier.
f(x) = 8 - 5x
f(4p+3) = 8 - 5(4p+3)
f(4p+3) = 8 - 20p - 15
f(4p+3) = -7-20p
This is the same as writing -20p - 7.
Answer:
A.
Step-by-step explanation:
When you substitute the coordinates to its corresponding variables, you can compare that answer choice A. is accurate compared to the other answer choices.
