<u>Given:</u>
Polynomial = P(x) = 3x² + kx + 6
Factor of the ablove polynomial = (x+3)
<u>To </u><u>Find</u><u>:</u>
Value of K, for which (x+3) become the factor of P(x) = 3x² + kx + 6
<u>Solu</u><u>tion</u><u> </u><u>:</u>
Now,
x + 3 = 0
⇒x = (-3)
So,
As (x+3) is a factor so x = (-3) is one root of the polynomial.
Therefore,
P(-3) = 0
→ P(-3) = 3(-3)² + k(-3) + 6 = 0
→ 3(9) - 3k + 6 = 0
→ 27 - 3k + 6 = 0
→ 27 + 6 - 3k = 0
→ 33 - 3k = 0
→ - 3k = -33
→ k = -33 ÷ -3
→ k = 11
- <u>Hence,For the value of k = 11, (x+3) is a factor of 3x²+ kx + 6</u>
<h2><u>V</u><u> </u><u>E</u><u> </u><u>R</u><u> </u><u>I</u><u> </u><u>F</u><u> </u><u>I</u><u> </u><u>C</u><u> </u><u>A</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h2>
3x²+ kx + 6, by putting the value of k = 11 and taking -3 as root the remainder should be zero
→ 3x²+ 11x + 6
→ 3(-3)² + 11(-3) + 6
→ 3(9) - 33 + 6
→ 27 - 33 + 6
→ 27 + 6 - 33
→ 33 - 33
→ 0
<u>Hence verified</u><u> </u><u>!</u>
Answer:
The answers are 0, 1 and −2.
Step-by-step explanation:
Let α=arctan(2tan2x) and β=arcsin(3sin2x5+4cos2x).
sinβ2tanβ21+tan2β2tanβ21+tan2β23tanxtan2β2−(9+tan2x)tanβ2+3tanx(3tanβ2−tanx)(tanβ2tanx−3)tanβ2=3sin2x5+4cos2x=3(2tanx1+tan2x)5+4(1−tan2x1+tan2x)=3tanx9+tan2x=0=0=13tanxor3tanx
Note that x=α−12β.
tanx=tan(α−12β)=tanα−tanβ21+tanαtanβ2=2tan2x−13tanx1+2tan2x(13tanx)or2tan2x−3tanx1+2tan2x(3tanx)=tanx(6tanx−1)3+2tan3xor2tan3x−3tanx(1+6tanx)
So we have tanx=0, tan3x−3tanx+2=0 or 4tan3x+tan2x+3=0.
Solving, we have tanx=0, 1, −1 or −2.
Note that −1 should be rejected.
tanx=−1 is corresponding to tanβ2=3tanx. So tanβ2=−3, which is impossible as β∈[−π2,π2].
The answers are 0, 1 and −2.
Answer:
Two points no intersections
Step-by-step explanation:
Answer:
6.25 litres
Step-by-step explanation:
ratio 1:4
25/4 = 6.25
4 * 6.25 = 25
1 * 6.25 = 6.25
= 6.25 litres of concentrate