Yousef typed a 36 word paragraph in 2/3 minute. what is his typing speed in words per minute
2 answers:
Answer:
Typing speed of Yousuf is 54 words per minute.
Step-by-step explanation:
Yousef typed a 36 word paragraph in minute.
Typing speed of Yousef will be given by the expression
Speed =
Now plug in the values in the expression
Typing speed =
=
= 54 words per minute
Therefore, typing speed of Yousuf is 54 words per minute.
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Given:
The equation of a line is
To find:
The slope and y-intercept of the line.
Solution:
We have,
...(i)
The slope intercept form of a line is
...(ii)
Where, m is the slope and b is the y-intercept.
On comparing (i) and (ii), we get
Slope of the line = 1
y-intercept of the line = -4
The coordinate form of y-intercept is (0,b). So, the y-intercept of the given line is (0,-4).
Therefore, the slope of the line is 1 and the y-intercept is (0,-4).
First find the secant line. The slope of the secant line through 
(when 
) and 
(when 
) is the average rate of change of 
over the interval ![[-1,2]](https://tex.z-dn.net/?f=%5B-1%2C2%5D)
:
The tangent line to will have a slope determined by the derivative:
Both the secant and tangent will have the same slope when
, or when 
.
The tangent line to
Both the secant and tangent will have the same slope when
Starting off with the polynomial in standard form would be extremely difficult, but we can construct one fairly easily with the zeroes we've been given.
We know from the given zeroes that our function has the value 0 when x = 1, x = -2, and x = 2. Manipulating each equation, we can rewrite them as x - 1 = 0, x + 2 = 0, and x - 2 = 0. To construct our polynomial, we simply use all three of the expressions on the left side of the equation as factors and multiply them together, obtaining:
Notice that we can easily obtain each our three zeroes by dividing both sides by the two other factors. From here, we just need to expand the left-hand side of the equation. I'll show the work required here:
=0\\ (x^2-x+2x-2)(x-2)=0\\ (x^2+x-2)(x-2)=0\\ (x^2+x-2)x-(x^2+x-2)2=0\\ x^3+x^2-2x-(2x^2+2x-4)=0\\ x^3+x^2-2x-2x^2-2x+4=0\\ x^3+(x^2-2x^2)+(-2x-2x)+4=0\\ x^3-x^2-4x+4=0\\](https://tex.z-dn.net/?f=%28x-1%29%28x%2B2%29%28x-2%29%3D0%5C%5C%0A%5Cbig%5B%28x-1%29x%2B%28x-1%292%5Cbig%5D%28x-2%29%3D0%5C%5C%0A%28x%5E2-x%2B2x-2%29%28x-2%29%3D0%5C%5C%0A%28x%5E2%2Bx-2%29%28x-2%29%3D0%5C%5C%0A%28x%5E2%2Bx-2%29x-%28x%5E2%2Bx-2%292%3D0%5C%5C%0Ax%5E3%2Bx%5E2-2x-%282x%5E2%2B2x-4%29%3D0%5C%5C%0Ax%5E3%2Bx%5E2-2x-2x%5E2-2x%2B4%3D0%5C%5C%0Ax%5E3%2B%28x%5E2-2x%5E2%29%2B%28-2x-2x%29%2B4%3D0%5C%5C%0Ax%5E3-x%5E2-4x%2B4%3D0%5C%5C)
So, in standard form, our cubic polynomial would be
We know from the given zeroes that our function has the value 0 when x = 1, x = -2, and x = 2. Manipulating each equation, we can rewrite them as x - 1 = 0, x + 2 = 0, and x - 2 = 0. To construct our polynomial, we simply use all three of the expressions on the left side of the equation as factors and multiply them together, obtaining:
Notice that we can easily obtain each our three zeroes by dividing both sides by the two other factors. From here, we just need to expand the left-hand side of the equation. I'll show the work required here:
So, in standard form, our cubic polynomial would be