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dlinn [17]
3 years ago
13

The wheels on Karen’s rollerblades each have a diameter of 5 cm. How far will the rollerblade travel in one turn of the wheel? U

se a calculator and round your answer to the nearest tenth of a centimeter.
Mathematics
1 answer:
Ierofanga [76]3 years ago
8 0
To find the distance in one turn of the wheel you have to find the circumference of the wheel.

Circumference = pi x d
= 3.14 x 5
= 15.7 cm

therefore the rollerblade wheel will travel 15.7 cm per rotation

hope this helps! if you did not understand a step or concept please let me know!
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puteri [66]
It is an open circle on 2 and then pointing to the left
4 0
3 years ago
Find the distance between the two points rounding to the nearest tenth (if
Sunny_sXe [5.5K]
Distance: (sqroot(x2-x1)^2 + (y2-y1)^2)
sqroot(1-3)^2 + (8-6)^2
sqroot(-2)^2 + (2)^2
sqroot(4) + 4
Square root of 8 = 2.82843
Round to nearest tenth
Solution: 2.8
7 0
3 years ago
The functions q and r are defined as follows
kolezko [41]

Answer:

98

Step-by-step explanation:

So we have the two functions:

g(x)=-2x-2\text{ and } r(x)=x^2-2

And we want to find the value of r(g(4)).

To do so, first find the value of g(4):

g(x)=-2x-2\\g(4)=-2(4)-2\\

Multiply:

g(4)=(-8)-2

Subtract:

g(4)=-10

Now, substitute this into r(g(4)):

r(g(4))\\=r(-10)

And substitute this value into r(x):

r(-10)=(-10)^2-2

Square:

r(-10)=100-2

Subtract:

r(-10)=98

Therefore:

r(-10)=r(g(4))=98

3 0
3 years ago
Describe in words how you would solve<br><br>the linear system y = 3x + 1 and y = - 2x + 3.
disa [49]

Answer:

Below.

Step-by-step explanation:

As both the right sides of the 2 equations are equal to y, by the transitive law of equality 3x + 1 = -2x + 3.

W then solve this equation for x then substitute this value of x  in the first equation ( y = 3x + 1) to find the value of y.  

3 0
3 years ago
What is the quotient 4-x/x^2+5x-6 divided by x^2-11x+28/x^2+7x+6 in simplified form?
sashaice [31]
\dfrac{4-x}{x^2+5x-6}:\dfrac{x^2-11x+28}{x^2+7x+6}=(*)

4-x=-(x-4)\\\\x^2+5x-6=x^2+6x-x-6=x(x+6)-1(x+6)=(x+6)(x-1)\\\\x^2-11x+28=x^2-7x-4x+28=x(x-7)-4(x-7)\\=(x-7)(x-4)\\\\x^2+7x+6=x^2+6x+x+6=x(x+6)+1(x+6)=(x+6)(x+1)\\\\(*)=\dfrac{-(x-4)}{(x+6)(x-1)}:\dfrac{(x-7)(x-4)}{(x+6)(x+1)}
=\dfrac{-(x-4)}{(x+6)(x-1)}\cdot\dfrac{(x+6)(x+1)}{(x-7)(x-4)}=\dfrac{-1}{x-1}\cdot\dfrac{x+1}{x-7}\\\\=-\dfrac{x+1}{x^2-7x-x+7}=-\dfrac{x+1}{x^2-8x+7}
7 0
3 years ago
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