Solve the equation A=B+c/d for C in terms of A,B and D.
1 answer:
You might be interested in
Answer:
in steps
Step-by-step explanation:
DE // BC
m∠ADE = m∠ABC and m∠AED = m∠ACB
∴ ΔADE similar to ΔABC
AB/AD = AC/AE
(AD + DB) / AD = (AE + EC) / AE
AD/AD + DB/AD = AE/AE + EC/AE
1 + DB/AD = 1 + EC/AE
DB/AD = EC/AE (AD/DB = AE/EC)
Answer:
x^4 -53x^2 +108x +160
Step-by-step explanation:
If <em>a</em> is a zero, then (<em>x-a</em>) is a factor. For the given zeros, the factors are ...
p(x) = (x +8)(x +1)(x -4)(x -5)
Multiplying these out gives the polynomial in standard form.
= (x^2 +9x +8)(x^2 -9x +20)
We note that these factors have a sum and difference with the same pair of values, x^2 and 9x. We can use the special form for the product of these to simplify our working out.
= (x^2 +9x)(x^2 -9x) +20(x^2 +9x) +8(x^2 -9x) +8(20)
= x^4 -81x^2 +20x^2 +180x +8x^2 -72x +160
p(x) = x^4 -53x^2 +108x +160
_____
The graph shows this polynomial has the required zeros.
