I need help asap<br><br>
what is the mode of 63, 70, 75, 80, 85, 85, 85, 92, 92, 95, 99
Ann [662]
The mode would be 85 since it occurs the most out of the whole data set.
The equation of the line that is parallel and passes through the point (2,3) is; x +2y =8.
<h3>What is the equation of the line that passes through (2,3)?</h3>
If follows from the task content that the slope of the first line is; (-4-0)/(4-(-4)) = -1/2.
Hence, since the lines are parallel, they have the same slope and the equation of the second line is;
-1/2 = (y-3)/(x-2)
-x+2 = 2y -6
x +2y =8.
Read more on equation of a line;
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Answer:
2.44
Step-by-step explanation:
Given: x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6
Also, R1 + R2 = 6
in order to find the value of a:
Let p(x) = x³ + 2x² - 5ax - 7 and q(x) = x³ + ax² - 12x + 6
Using remainder theorem i.e if a polynomial p(x) is divisible by polynomial of form x - a then remainder is given by p(a).
Then,
R1 = p( -1 ) = (-1)³ + 2(-1)² - 5a(-1) - 7 = -1 + 2 + 5a - 7 = 5a - 6
R2 = q( 2 ) = 2³ + a(2)² - 12(2) + 6 = 8 + 4a - 24 + 6 = 4a - 10
Now,
R1 + R2 = 6
5a - 6 + 4a - 10 = 6
9a = 22
a=2.44
Therefore, Value of a is 2.44
Answer:
80 m^2
Step-by-step explanation:
The given information lets you write two equations involving length (x) and width (y).
- 2(x +y) = 36 . . . . the perimeter is 36 m
- (x+1)(y+2) -xy = 30 . . . . increasing the length and width increases area
The second of these equations simplifies to another linear equation, giving a system of linear equations easily solved.
xy +y +2x + 2 -xy = 30
2x +y = 28 . . . . . . . subtract 2
Dividing the first equation by 2 gives
x +y = 18
and subtracting this from the above equation gives ...
(2x +y) -(x +y) = 28 -18
x = 10
Then
y = 18 -10 = 8
The area of the original rectangle is xy = 10·8 = 80 m^2.
Answer:
D.
Step-by-step explanation:
Remember that the limit definition of a derivative at a point is:
![\displaystyle{\frac{d}{dx}[f(a)]= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28a%29%5D%3D%20%5Clim_%7Bx%20%5Cto%20a%7D%5Cfrac%7Bf%28x%29-f%28a%29%7D%7Bx-a%7D%7D)
Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:
![\displaystyle{\frac{d}{dx}[f(1)]= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%281%29%5D%3D%20%5Clim_%7Bx%20%5Cto%201%7D%5Cfrac%7B%5Cln%28x%2B1%29-%5Cln%282%29%7D%7Bx-1%7D%7D)
Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).
The answer will thus be D.
