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3 years ago

VER IMPORTANT NEED HELP ASAP Is a hotdog... a sandwich?

Mathematics

2 answers:

AleksAgata [21]3 years ago

8 0

here's the real question is cereal a soup

Alex17521 [72]3 years ago

7 0

Answer:

Step-by-step explanation:

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Drag the tiles to the boxes to form correct pairs.<br> Match the pairs of equivalent expressions.

Masja [62]

Answer:

Step-by-step explanation:

look this solution

5 0

3 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:

#learnwithBrainly

5 0

3 years ago

Brook makes this conjecture: Because the angles are equal, the proportion of area of each sector relative to the area of the giv

AnnZ [28]

Answer:

The Brooks conjecture is true as the proportion of the area of a sector to the area of a circle is only dependent on the angle.

Step-by-step explanation:

The area of a sector for a radius r and an angle θ is given as

$\Delta Sector=\dfrac{1}{2}r^2\theta$

Whereas the area of a circle for a radius r is given as

$\Delta Circle=\pi r^2$

Now the ratio is given as

$\dfrac{\Delta Sector}{\Delta Circle}=\dfrac{\dfrac{1}{2}r^2\theta}{\pi r^2}$

Here as r is the radius of the circle and it is the same therefore the above ratio is simplified to

$\dfrac{\Delta Sector}{\Delta Circle}=\dfrac{\theta}{2\pi }$

Here as π is constant the only variable here is θ. Thus according to Brooks conjecture when the angles are equal, the proportion of area of each sector relative to the area of the given circle is also equal is true.

6 0

3 years ago

5x - 10 + 50 = 9x - 8<br> Solve for x

alexandr402 [8]

Order them in which the number with x are on one side and the rest are on the other side
5x-9x=-8+10-50
-4x=-48
X=-48/-4
X=12

4 0

3 years ago

When Shanaje baby-sits her nephews, she charges $8 plus $4 per hour. If

Vlad1618 [11]

4h +8=20
Hours worked times rate paid plus 8 must equal 20

3 0

3 years ago