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dalvyx [7]
3 years ago
11

The line plots shows the distance, in miles, that students walked on Friday. How many students walked over 1/2 of a mile.

Mathematics
1 answer:
otez555 [7]3 years ago
3 0
U count how many students were walking then multiply it by half itself
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Form a quadratic polynomial whose zeroes are 3-√3/5 and 3+√3/5?​
barxatty [35]

Answer:

x^2 -6x + 222/25

Step-by-step explanation:

If the zeros are as above, then ;

x = 3-√3/5 or x = 3 + √3/5

Firstly, let’s represent √3/5 by b

Thus;

The two roots are ;

x = 3-b or x = 3 + b

so;

x+ b -3 and x -3-b

The quadratic equation is the product of the two

(x + b-3)(x - b -3)

x(x - b-3) + b(x -b -3) -3(x - b -3)

= x^2 -bx -3x + bx -b^2 -3b -3x + 3b + 9

Collect like terms and we are left with;

x^2 -6x -b^2 + 9

So let’s put back b = √3/5

x^2 -6x -(√3/5)^2 + 9

x^2 -6x -3/25 + 9

x^2 -6x + 222/25

3 0
3 years ago
The quotient if a number and 9 times it’s reciprocal is 1. Find the number. (Might be more than one answer)
astra-53 [7]
You need to add 9 and 1 which gives you 10
5 0
4 years ago
Read 2 more answers
Which statements about acceleration are true?
Rom4ik [11]
Is there a graph or something?
4 0
3 years ago
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WHat are the correct answers to this?
iogann1982 [59]

Answer:

1st blank _ 36

2nd blank _ -30

last blank _ 6

7 0
2 years ago
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The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

5 0
3 years ago
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