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Komok [63]
2 years ago
13

Is the dilation from A to B a reduction or enlargement?

Mathematics
2 answers:
WITCHER [35]2 years ago
8 0

Answer:

This is actually a reduction, and the scale factor of this dilation is 0.5.

Masteriza [31]2 years ago
4 0
It’s a because the factor reduce by itself
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If f(x)=x^3-x, what is the average rate of change of f(x) over the interval [1,5]?
RSB [31]

Answer:

30

Step-by-step explanation:

Average rate of change=Δf(x)/Δx = [f(x2)-f(x1]/[x2-x1]

Δf(x)/Δx = [f(5)-f(1]/[5-1]

f(5)=5³-5=120

f(1)=1³-1=0

Δf(x)/Δx = [120-0]/[5-1]=120/4=30

4 0
3 years ago
Can someone please answer this question please answer it correctly and lease show work please help me I need it
earnstyle [38]

Answer:

C

Step-by-step explanation:

The correct answer would be \frac{-5}{3}.  Therefore, none of the answer choices fulfill this and A and B are not correct.

8 0
3 years ago
Read 2 more answers
How is the product of 2 and -5 shown on a number line
otez555 [7]

Answer: on point -10

Step-by-step explanation:

Because 2*-5 is -10.

8 0
3 years ago
Read 2 more answers
An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
n a survey about favorite fruits, 28 people chose apples, 17 chose oranges, and the rest chose pears. If 60 people were surveyed
nordsb [41]

Based on the number of people who chose the other fruits, the percentage of people who chose pears was <u>25%. </u>

<h3>Number of people who chose pears </h3>

= Number of participants - Number who chose apples - Number who chose oranges

= 60 - 28 - 17

= 15 people

<h3>Percentage who chose pears </h3>

= Number of people who chose pears / Total number of people x 100%

= 15 / 60 x 100%

= 25%

In conclusion, 25% chose pears.

Find out more on percentages at brainly.com/question/13516343.

7 0
2 years ago
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