Answer:
<em>f(x)=x²-3x-10</em>
Step-by-step explanation:
\begin{gathered}f(x) = x {}^{2} - 3x - 10 \\ to \: find \: x \: intercept \:o r \: zero \: substitute \: f(x) = 0\: \\ 0 = x {}^{2} - 3x - 10 \\ x {}^{2} - 3x - 10 = 0 \\ x {}^{2} + 2x - 5x - 10 = 0 \\ x(x + 2) - 5x - 10 = 0 \\ x(x + 2) - 5(x + 2) = 0 \\ (x + 2).(x - 5) = 0 \\ x + 2 = 0 \\ x - 5 = 0 \\ x = - 2 \\ x = 5\end{gathered}
f(x)=x
2
−3x−10
tofindxinterceptorzerosubstitutef(x)=0
0=x
2
−3x−10
x
2
−3x−10=0
x
2
+2x−5x−10=0
x(x+2)−5x−10=0
x(x+2)−5(x+2)=0
(x+2).(x−5)=0
x+2=0
x−5=0
x=−2
x=5
therefore the zeros of the equation are x₁=-2,x₂=5
Answer:
5%
Step-by-step explanation:
487/9374=0.05
0.05 would translate to 5%
Hope this helps! :)
Answer:
At 25 = 6.8612mm
At 50 years = 5.422mm
Step-by-step explanation:
Equation,
d = 2.115Logₑa + 13.669
d = diameter of the pupil
a = number of years
Note : Logₑa = In a (check logarithmic rule)
d = 2.115Ina + 13.669
1. At 25 years,
d = -2.115In25 + 13.669
d = -2.115 × 3.2188 + 13.669
d = -6.807762 + 13.669
d = 6.8612mm
At 25 years, the pupil shrinks by 6.86mm
2. At 50 years,
d = -2.1158In50 + 13.669
d = -2.1158 * 3.912 + 13.669
d = -8.2770 + 13.699
d = 5.422mm
At 50 years, the pupil shirks by 5.422mm
To save this question, I had to plug in the values into the equation.
Solving for Logₑa might be difficult, so instead I used Inx which is the same thing. Afterwards, i substituted in the values and solve the equation for each years.
Answer:
<h2> <em><u>x = 2.82 (approx.)</u></em></h2>
Step-by-step explanation:
<em><u>Given</u></em><em><u>, </u></em>
- 356 = -77 - 99x
=> -77 - 99x = -356
=> -99x = -356 + 77
=> -99x = -279
=> x = 2.8181....
=> <em><u>x = 2.82 (approx.) (Ans)</u></em>
PV = nRT.......divide both sides by nR
(PV) / (nR) = T
