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3 years ago

Subtract and simplify.

Mathematics

2 answers:

slamgirl [31]3 years ago

8 0

It is the last option the
4y^2-10y-1

Anestetic [448]3 years ago

6 0

$(y^2-7y-5)-(-3y^2+3y-4)$
$=y^2-7y-5+3y^2-3y+4$
$=3y^2+y^2-7y-3y-5+4$
$=4y^2-10y-1$

In short, the answer would be $4y^2-10y-1$ (Option D).

Hope this helps !

Photon

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

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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

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4 years ago

In triangle JKL, sin(b°) = three fifths and cos(b°) = four fifths. If triangle JKL is dilated by a scale factor of 2, what is ta

scoundrel [369]

Answer:

Tan(b°) = 3/4 = three fourths (C)

Step-by-step explanation:

Triangle JKL is a right angled triangle in which angle K is a right angle and angle L= b°

We would apply SOHCAHTOA from trigonometry to determine the value for each sides.

For ∆JKL

Sin(b°) = opposite/hypotenuse

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Cos(b°) = adjacent/hypotenuse

Cos(b°) = 4/5

Tan(b°) = 3/4

From the above we know the value of each sides of ∆JKL

Find attached the diagram of the above triangle (1)

∆JKL is dilated by a scale factor = 2

Meaning we would multiply each sides of ∆JKL by 2. The values of the new triangle become:

Opposite = 2(3) = 6

Adjacent = 2(4) = 8

Hypotenuse = 2(5) = 10

To find tan(b°) of the new triangle, we would apply tangent ratio

Tan(b°) = opposite/adjacent

Tan(b°) = 6/8

Tan(b°) = 3/4

Find attached the diagram for the new triangle (2)

Diagram 3 shows the drawing of both triangles together.

From the above, we can see the angle doesn't change when a shape is dilated by a scale factor.

Therefore, tan(b°) = three fourths (C)

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