T(1) = 3, t(n) = -2t(n-1) + 1
So t(2) = -2(t(1)) + 1 = -2(3) + 1 = -5
t(3) = -2(t(2)) +1 = -2(-5) +1 = 11
t(4) = -2(t(3)) +1 = -2(11) +1 = -21
t(5) = -2(t(4)) +1 = -2(-21) +1 = 43
Answer:
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Find a point-slope form for the line that satisfies the stated conditions. Slope , passing through (-5,4)
I really need this question answered
By:
I don't see a value for the slope. We need that to set the equation, otherwise I can write an unlimited number of equations that pass through (-5,4).
I'll assume a slope so that you can see how the procedure would work. I like 6, so we'll assume a slope of 6.
The equation for a straight line has the form y = mx + b, where m is the slope and y is the y-intercept, the value of y when x = 0. We want a line that has slope 6, so:
y = 6x + b
We need to find b, so substitute the point (-5,4) that we know is on the line:
4 = 6*(-5) + b and solve for b
4 = -30 + b
b = 34
The line is y = 6x + 34
Answer:
0.81 & 0.83
Step-by-step explanation:
Answer:
y²+5y+4=y²+4y+y+4=y(y+4)+1(y+4)=
<u>(y+4)(y+1)</u>
Answer:
Line segment TS.
Step-by-step explanation:
We have an Unit Circle. The y-coordinate of T is the sine of 
. Therefore, the exact value of 
is TS.
