Answer:
First, find tan A and tan B.
cosA=35 --> sin2A=1−925=1625 --> cosA=±45
cosA=45 because A is in Quadrant I
tanA=sinAcosA=(45)(53)=43.
sinB=513 --> cos2B=1−25169=144169 --> sinB=±1213.
sinB=1213 because B is in Quadrant I
tanB=sinBcosB=(513)(1312)=512
Apply the trig identity:
tan(A−B)=tanA−tanB1−tanA.tanB
tanA−tanB=43−512=1112
(1−tanA.tanB)=1−2036=1636=49
tan(A−B)=(1112)(94)=3316
kamina op bolte
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You can solve this by using "similar triangles".
In triangle ABC, we are looking for side AC which is x. Side AC is similar to side DF in triangle EDF.
You can solve for side x by picking two sides in triangle ABC and their corresponding sides in triangle EDF. This is what I mean:

Substitute for the values of AC, BC, DF and EF:


To solve for y, do the same thing. Pick two sides on triangle ABC and their corresponding sides in triangle DEF.

Substitute for the values and solve:


We have the value x to be 5.5 units and y to be 6 units.
Answer:
hi
Step-by-step explanation:
Answer:
<h2>-33x - 82</h2>
Step-by-step explanation:
Use
PEMDAS:
P Parentheses first
E Exponents (ie Powers and Square Roots, etc.)
MD Multiplication and Division (left-to-right)
AS Addition and Subtraction (left-to-right)
and the distributive property a(b + c) = ab + ac.
![1)\qquad5(x+3)=5x+15\\\\2)\qquad x+5(x+3)=x+5x+15=6x+15\\\\3)\qquad2[x+5(x+3)]=2(6x+15)=12x+30\\\\4)\qquad x-2[x+5(x+3)]=x-(12x+30)=x-12x-30=-11x-30\\\\5)\qquad3\{x-2[x+5(x+3)]\}=3(-11x-30)=-33x-90\\\\6)\qquad8+3\{x-2[x+5(x+3)]\}=8+(-33x-90)=-33x-82](https://tex.z-dn.net/?f=1%29%5Cqquad5%28x%2B3%29%3D5x%2B15%5C%5C%5C%5C2%29%5Cqquad%20x%2B5%28x%2B3%29%3Dx%2B5x%2B15%3D6x%2B15%5C%5C%5C%5C3%29%5Cqquad2%5Bx%2B5%28x%2B3%29%5D%3D2%286x%2B15%29%3D12x%2B30%5C%5C%5C%5C4%29%5Cqquad%20x-2%5Bx%2B5%28x%2B3%29%5D%3Dx-%2812x%2B30%29%3Dx-12x-30%3D-11x-30%5C%5C%5C%5C5%29%5Cqquad3%5C%7Bx-2%5Bx%2B5%28x%2B3%29%5D%5C%7D%3D3%28-11x-30%29%3D-33x-90%5C%5C%5C%5C6%29%5Cqquad8%2B3%5C%7Bx-2%5Bx%2B5%28x%2B3%29%5D%5C%7D%3D8%2B%28-33x-90%29%3D-33x-82)