Answer:
B. y = 2 / 3 x + 1
Step-by-step explanation:
Looking at the graph, you can see that the y-intercept is 1, so you can eliminate choices A and D. You are now left between choices B and C. To figure out the slope, you need to do rise and run. From the y-intercept, you rise up 2, and you run right 3. This means that the slope is 2/3.
Answer:
Step-by-step explanation:
1) ?
I can't see what that one given angle is .. but it's that mystery angle subtracted from 180 give the angle BOA then you know that's an isosceles triangle again that the BOA is part of... so then just subtract BOA from 180 to find the two other angle of that triangle... they are the small so just divide your answer of 180-BOA /2 is the angle of each.. then since you know those to smaller angles subtract one from 90 to find the angle BAX
2) ( as we would read normally)
You're making this really tough on me.. I can just barely read the equations
I think it's 1+6x and 7x-3 . b/c they are the same length sides you can set those equal
1+6x = 7x -3
4+6x = 7x
4 = x
that worked out well :
for the tangent.. it's Tan(Ф)= Opp/ Adj
but I don't know which side they want to solve for.. I think you may have left off some of the instructions???? :/
ohh I think they really mean.. what's the length of the tangent lines .. that was confusing to me.. :/ just plug in 4 into x for either eq.
1+6(4) = 25
or
7(4)-3=25
tangent is 25
3)
x-2 = 2x-10
x+8 = 2x
8 = x
again they made it work out easy :)
Then plug 8 into either equation to find the length of the tangent lines
8-2=10
tangent is 10
4) = 2) ??? they are the same question maybe you meant to put something else?
(3)
It says the rocket was in the air for approximately 6 seconds before hitting the ground.
But the graph proves it untrue because after 6 seconds it is still in the air, rather than on the ground like (3) suggests.
