(2x^2+3x+4)(x^3+4x^2+3x)(x+4)
First use the distributive property for the first two:
(2x^5+8x^4+6x^3+3x^4+12x^3+9^2+4x^3+16x^2+12x)(x+4)
And then combine like terms:
(2x^5+11x^4+22x^3+25x^2+12x)(x+4)
Continue use the distributive property:
(2x^6+11x^5+22x^4+25x^3+12x^2+8x^5+44x^4+88x^3+100x^2+48x)
Continue combine like terms:
(2x^6+19x^5+66x^4+113x^3+112x^2+48x)
And that is your final answer.
Hope that help:)
The number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
- Since one student runs at a speed of 180 feet per minutes, in time t minutes, he moves a distance of d = 180t.
- Also, the second student runs at a speed of 160 feet per minute, in time t minutes, he moves a distance of d' = 160t.
Since they are initially 10 feet apart, their total distance apart after t minutes is D = d + 10 + d'
D = 180t + 10 + 160t
D = 340t + 10
<h3>Number of minutes before they are 1870 feet</h3>
Making t subject of the formula, we have
t = (D - 10)/340
Since they are 1870 feet apart after t minutes, D = 1870 feet.
t = (D - 10)/340
t = (1870 - 10)/340
t = 1860/340
t = 5.47 minutes
So, the number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
Learn more about minutes of distance apart here
brainly.com/question/8783264
Answer:
Step-by-step explanation:
a. P (red) = <u>2/9</u>
b. P (blue) = 3/9 = <u>1/3</u>
c. P (green) = <u>4/9</u>
d. Green, because out of the three types of gumballs, the green gumballs are more in number, therefore, it is the one you are most likely to get.
Answer:
μ = 0.169
ME = 0.051
Step-by-step explanation:
The confidence interval is:
CI = μ ± ME
So the mean is the middle of the confidence interval, and the margin of error is half the difference.
μ = (0.118 + 0.220) / 2 = 0.169
ME = (0.220 − 0.118) / 2 = 0.051
The given figure shows a vertical hyperbola with its centre at origin, and as we observe the figure, we can conclude that :
Length of transverse axis is :
length of conjugate axis is :
Equation of hyperbola ~
plug in the values ~
