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3 years ago

A store is offering 20%

Mathematics

1 answer:

Furkat [3]3 years ago

4 0

Answer: Please expand the question

Step-by-step explanation:

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G evaluate lim h → 0 (1 + h)9 − 1 h . hint: this limit represents the derivative of a function f at a given point

Bond [772]

$\displaystyle\lim_{h\to0}\frac{(1+h)^9-1}h$

Write $1=1^9$ , and recall that for a differentiable function $f(x)$ , the derivative at a point $x=a$ is given by

$f'(a)=\displaystyle\lim_{h\to0}\frac{f(a+h)-f(a)}h$

which would suggest that for this limit, $f(x)=x^9$ and $a=1$ . We have $f'(x)=9x^8$ , and so the value of the limit would be $9(1)^8=9$ .

7 0

3 years ago

Consider the equations: y = 5x + 2 and y = 3x + 4.

LenKa [72]

Answer:

Step-by-step explanation:

The answer is:

x=1

y=7

7 0

3 years ago

Describe how to use a tens fact to find the difference for 15-8.

Papessa [141]

I subtracted 15-8 gave me 7 then I took
10- my answer witch was 7
10-7=3

7 0

4 years ago

Andrew is pouring a square shaped patio and needs 529 square feet of concrete what is the length of each side of the patio? Expl

LekaFEV [45]

Answer:

The answer is 23 feet on each side

Step-by-step explanation:

Sqrt of 529= 23

Hope it works

4 0

3 years ago

Determine which function has the greatest rate of change over the interval [0, 2].

ruslelena [56]

Remember that the average rate of change of a function over an interval is the slope of the straight line connecting the end points of the interval. To find those slopes, we are going to use the slope formula: $m= \frac{y_{2}-y_{1}}{x_2-x_1}$

Rate of change of $a$ :
From the graph we can infer that the end points are (0,1) and (2,4). So lets use our slope formula to find the rate of change of $a$ :
$m= \frac{y_{2}-y_{1}}{x_2-x_1}$
$m= \frac{4-1}{2-0}$
$m= \frac{3}{2}$
$m=1.5$
The average rate of change of the function $a$ over the interval [0,2] is 1.5

Rate of change of $b$ :
Here the end points are (0,0) and (2,2)
$m= \frac{2-0}{2-0}$
$m= \frac{2}{2}$
$m=1$
The average rate of change of the function $b$ over the interval [0,2] is 1

Rate of change of $c$ :
Here the end points are (0,-1) and (2,0)
$m= \frac{0-(-1)}{2-0}$
$m= \frac{1}{2}$
$m=0.5$
The average rate of change of the function $c$ over the interval [0,2] is 0.5

Rate of change of $d$ :
Here the end points are (0,0.5) and (2,2.5)
$m= \frac{2.5-0.5}{2-0}$
$m= \frac{2}{2}$
$m=1$
The average rate of change of the function $d$ over the interval [0,2] is 1

We can conclude that the <span>function that has the greatest rate of change over the interval [0, 2] is the function a.</span>

4 0

4 years ago