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3 years ago

Midpoint: (3,5) and (-1, 7)

Mathematics

1 answer:

8_murik_8 [283]3 years ago

8 0

To find the midpoint:
x=3-1/2= 1
y=5+7/2= 6

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A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s. (a) What is the average velocity of the

Archy [21]

Answer:

<h2>a) Average velocity = 1.25 m/s to the right</h2><h2>b) Average speed = 3.75 m/s</h2>

Step-by-step explanation:

A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s.

Total displacement = 20 - 10 = 10 m to the right

Total time = 5+3 = 8 s

Total distance = 20 + 10 = 30 m

a) We have

Total displacement = Average velocity x Time

10 = Average velocity x 8

Average velocity = 1.25 m/s to the right

b)We have

Total distance = Average speed x Time

30 = Average speed x 8

Average speed = 3.75 m/s

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3 years ago

F (x) = 2x^2 - 4x + 1

Kobotan [32]

Answer:

<h2><u>tinanong ko rin po yang tanong nyo at ito po ang sagot nila</u></h2>

3 0

2 years ago

32 minus twice p is 31

Katena32 [7]

P=1 because 32-1 is 31

3 0

3 years ago

At what values of x does f(x) = x^3 - 2x^2 -4x+1 satisfy the mean value theorwm on [0,1]

denis-greek [22]

Answer:

x=1/3

Step-by-step explanation:

A function f is given as

$f(x) = x^3-2x^2-4x+1$ in the interval [0,1]

This function f being an algebraic polynomial is continuous in the interval [0,1] and also f is differntiable in the open interval (0,1)

Hence mean value theorem applies for f in the given interval

$f(1) = 1-2-4+1 = -4\\f(0) = 1$

The value

$\frac{f(1)-f(0)}{1-0} =\frac{-4-1}{1} =-5$

Find derivative for f

$f'(x) = 3x^2-4x-4$

Equate this to -5 to check mean value theorem

$3x^2-4x-4=-5\\3x^2-4x+1=0\\\\(x-1)(3x-1) =0\\x= 1/3 : x = 1$

We find that 1/3 lies inside the interval (0,1)

4 0

2 years ago

denis-greek [22]

Answer:

Part a) $AB=15\ cm$

Part b) $AB=11.1\ cm$

Part c) $AB=3/5\ cm$

Part d) $AB=3s\ cm$

Step-by-step explanation:

see the attached figure to better understand the question

we know that

To find the length of the image after a dilation, multiply the length of the pre-image by the scale factor

Part a) we have

The scale factor is 5

The length of the pre-image is 3 cm

therefore

The length of the image AB after dilation is

$3(5)=15\ cm$

Part b) we have

The scale factor is 3.7

The length of the pre-image is 3 cm

therefore

The length of the image AB after dilation is

$3(3.7)=11.1\ cm$

Part c) we have

The scale factor is 1/5

The length of the pre-image is 3 cm

therefore

The length of the image AB after dilation is

$3(1/5)=3/5\ cm$

Part d) we have

The scale factor is s

The length of the pre-image is 3 cm

therefore

The length of the image AB after dilation is

$3(s)=3s\ cm$

7 0

3 years ago