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2 years ago

In need of assistance

Mathematics

2 answers:

Mkey [24]2 years ago

7 0

Answer:

Step-by-step explanation:

_---&$'_-4$':/-&$$--(:'##4--&&4___&$$_666&4&&&&&

babunello [35]2 years ago

4 0

Answer: 489.8 in

Step-by-Step Explanation:

Diameter (d) = 26 in
Radius (r) = 26/2 = 13 in

No. Of Revolutions = 6

We know, that the length around the tire is nothing but the perimeter of the tire.

Perimeter = 2πr

Therefore,
= 2πr
= 2 * π * r
= 2 * 3.14 * 13
= 6.28 * 13
=> 81.64

Perimeter = 81.64 in

Distance Travelled after 6 Revolutions of the Tire = Perimeter * 6

Therefore,
= 81.64 * 6
=> 489.8

Distance Travelled after 6 Revolutions of the Tire = 489.8 in

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Write a rule for the function represented by the table.

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y=19x+9 would be the rule

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4 years ago

20 POINTS ASAP The graphed line can be expressed by which equation? Math item stem image y+3=32(x+1) y−0=−32(x+1) y−3=32(x−1) y−

storchak [24]

Answer:

The Answer is: y - 3 = 3/2(x - 1)

Step-by-step explanation:

Given Points: (1, 3) and (-3, -3)

Find the slope m:

m = y - y1 / (x - x1)

m = 3 - (-3) / (1 - (-3))

m = 3 + 3 / 1 + 3

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y - y1 = m(x - x1)

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4 years ago

Read 2 more answers

Which statement best reflects the solution(s) of the equation?

Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0$

Now, factoring the term:

$x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2$

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ the denominator becomes zero i.e

$\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}$

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2 in the equation,

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$

$\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2$

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

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4 years ago

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To convert from a fraction to a percent,
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3 years ago

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I am Lyosha [343]

Answer:

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Step-by-step explanation:

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