Answer: 1+p= 9 and 0.5l + 1.5p = 9.5
Step-by-step explanation:
Since y = 5x-1, we can fill it into 3x + 3y = -3. First, let's look at relating to a simpler equation. Let's say that x + y = 9 and y = 3 + 5. Now, we can fill it in to get that x + (3 + 5) = 9. Now, we know that 3+5 is 8, so x + 8 = 9. Now, x = 1. Likewise, we can do the same. For 3x + 3y = -3, all we need to do is to switch the y in 3x + 3y = -3 with 5x - 1. So it would become 3x + 3(5x - 1) = -3. Now we distribute to get 3x + 15x - 3 = -3. Now add three to both sides to get 3x + 15x = 0. Now simplify to get 18x = 0. Now we know that x = 0. Now fill x into y = 5x - 1. So y = 5(0) - 1. Now we know that y = -1.
To check fill in the answer to 3x + 3y = -3.
3(0) + 3(-1) = -3
0 + (-3) = -3
0 - 3 = -3
-3 = -3
Now that our check is completed we now know that x is 0 and y is -1.
Answer:
-7x
Step-by-step explanation:
==> -12x + 5x.
==> -12+5xx
==> -7x
Sin A = opp/hyp = 4/5
Cos A = adj/hyp = 3/5
Tan A = opp/adj = 4/3
Cos 27 = 0.8910065... --> 0.89
You can just plug that into your graphing calculator, just make sure the degree mode is on.
So lets get to the problem
<span>165°= 135° +30° </span>
<span>To make it easier I'm going to write the same thing like this </span>
<span>165°= 90° + 45°+30° </span>
<span>Sin165° </span>
<span>= Sin ( 90° + 45°+30° ) </span>
<span>= Cos( 45°+30° )..... (∵ Sin(90 + θ)=cosθ </span>
<span>= Cos45°Cos30° - Sin45°Sin30° </span>
<span>Cos165° </span>
<span>= Cos ( 90° + 45°+30° ) </span>
<span>= -Sin( 45°+30° )..... (∵Cos(90 + θ)=-Sinθ </span>
<span>= Sin45°Cos30° + Cos45°Sin30° </span>
<span>Tan165° </span>
<span>= Tan ( 90° + 45°+30° ) </span>
<span>= -Cot( 45°+30° )..... (∵Cot(90 + θ)=-Tanθ </span>
<span>= -1/tan(45°+30°) </span>
<span>= -[1-tan45°.Tan30°]/[tan45°+Tan30°] </span>
<span>Substitute the above values with the following... These should be memorized </span>
<span>Sin 30° = 1/2 </span>
<span>Cos 30° =[Sqrt(3)]/2 </span>
<span>Tan 30° = 1/[Sqrt(3)] </span>
<span>Sin45°=Cos45°=1/[Sqrt(2)] </span>
<span>Tan 45° = 1</span>
