Can someone help me, I don't need to see how u did it just make sure it is correct.
1 answer:
Y = - 4x^2 - 3x + 3 This thing is complex with fractions. I have to work it out.
y = -4(x^2 + 3/4 x) + 3
y = -4(x^2 + 3/4 x + (3/8)^2 ) + 3
y = -4(x + 3/8)^2 +3 + 4*(9/64)
y = -4(x + 3/8)^2 + 3 + 9/16
y = -4(x + 3/8)^2 + 3 9/16 Equation in Vertex form. <<<<< answer
y = -4(x + 0.385) + 3.5625 Equation in Vertex form. <<<< answer
One is in decimal form. One in fraction form.
the vertex (-3/8, 3 9/16) or alternately
the vertex (-0.375,3.5625)
You might want the graph to confirm my answer.
y = -4(x^2 + 3/4 x) + 3
y = -4(x^2 + 3/4 x + (3/8)^2 ) + 3
y = -4(x + 3/8)^2 +3 + 4*(9/64)
y = -4(x + 3/8)^2 + 3 + 9/16
y = -4(x + 3/8)^2 + 3 9/16 Equation in Vertex form. <<<<< answer
y = -4(x + 0.385) + 3.5625 Equation in Vertex form. <<<< answer
One is in decimal form. One in fraction form.
the vertex (-3/8, 3 9/16) or alternately
the vertex (-0.375,3.5625)
You might want the graph to confirm my answer.
You might be interested in
Answer:
The probability there will be between 280 and 360 column inches of classified advertisement
P(280≤X≤360) = P(-2≤Z≤2) = 0.9544
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean the population = 320
The standard deviation of the Population = 20
Let 'X' be the random variable in a normal distribution
Let 'X' = 280
Let 'X' = 360
<u><em>Step(iii):-</em></u>
The probability there will be between 280 and 360 column inches of classified advertisement
P(280≤X≤360) = P(-2≤Z≤2)
= P(z≤2) -P(z≤-2)
= P(z≤2)+P(z≤2)
= 2P(z≤2)
= 2×0.4772 ( from normal table)
= 0.9544