The general form of a quadratic equation is that which is expressed in a form,
ax² + bx + c = 0
To write the equation in its general form, eliminate the decimal by multiplying the equation by 10. It now becomes,
3x² + 50x - 70 = 0
The value of b is therefore 50 and the answer is the second choice.
Part a.
<span>The perimeter formula is: </span>
P = 2w + 2LYour answer for part A is
P = 2w + 2LPart b.
They tell us in part b, that:
w = 6L + 20
Substitute that into what we got for part a.
P = 2(6L + 20) + 2L
P = 12L + 40 + 2L
P = 14L + 40Your answer for partb is
P = 14L + 40Part c.
So we take what we got from part b, and solve for L.
P = 14L + 40
14L = P - 40

That is your answer for part 3 ^
part d.
Plug in w = 32 into what we first got in part b.
w = 6L + 20
32 = 6L + 20
6L = 12
L = 2Your answer for part d is
L = 2<span>:D</span>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>:</em><em>)</em><em>✌</em>
The objective is to state why the value of
converging alternating seies with terms that are non increasing in magnitude
lie between any two consecutive terms of partial sums.
Let alternating series
<span>Sn = partial sum of the series up to n terms</span>
{S2k} = sequence of partial sum of even terms
{S2k+1} = sequence of partial sum of odd terms
As the magnitude of the terms in the
alternating series are non-increasing in magnitude, sequence {S2k} is bounded
above by S1 and sequence {S2k+1} is bounded by S2. So, l lies between S1 and
S2.
In the similar war, if first two terms of the
series are deleted, then l lies in between S3 and S4 and so on.
Hence, the value of converging alternating
series with terms that are non-increasing in magnitude lies between any two
consecutive terms of partial sums. So, the remainder Rn = S – Sn alternating
sign
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