sum of sequence Find the sum of 46 + 42 + 38 + ... + (-446) + (-450) is -25,250
<u>Step-by-step explanation:</u>
We need to find sum of sequence : 46 + 42 + 38 + ... + (-446) + (-450)
Given sequence is an AP with following parameters as :
So , Let's calculate how many terms are there as :
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Sum of an AP is :
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Therefore , sum of sequence Find the sum of 46 + 42 + 38 + ... + (-446) + (-450) is -25,250
I´d say "d" is the distance from the eye to the wall.
Now substracting 1.2-1 you´ll get the distance of the wall of the smallest triangle = 0.2 And you do 1.5-0.2= 0.3 that´s the distance of the wall of the other triangle. Then you solve everything with Pitagoras theorem. You have 2 rectangle triangles.
B+alfa=45°
tan^-1(0.2/d)=B
tan^-1(1.3/d)=alfa
THEN:
tan^-1(0.2/d)+tan^-1(1.3/d)=45°
Now you have 3 ecs and 3 variables.
alfa,B and "d"
150.
You can read the uppermost point on the graph which is pretty much that point.
Answer:
11/61
Step-by-step explanation:
cos b = adj side/hypotenuse
=11/61