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3 years ago

What is the solution to the rational equation x over 2x plus 1 minus 1 over 4 equals 2 over 2x plus 1

1 answer:

5 0

We want to solve
$\frac{x}{2x+1} - \frac{1}{4} = \frac{2}{2x+1}$

Subtract 2/(2x+1) from each side.
$\frac{x}{2x+1} - \frac{2}{2x+1y} - \frac{1}{4} = 0$

Add 1/4 to each side, and simplify the left side.
$\frac{x-2}{2x+1} = \frac{1}{4}$

Cross multiply.
4(x - 2) = 1(2x + 1)
4x - 8 = 2x + 1

Simplify and solve for x.
4x - 2x = 1 + 8
2x = 9
x = 9/2 or x = 4 1/2.

Answer: $x = \frac{9}{2} \,\, or \,\, x = 4 \frac{1}{2}$

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Write an equation for the line parallel to the given line that contains C.<br> C(2,8); y = - 4x + 3

Luba_88 [7]

Answer: i love brinly its the best

the value of x is 4, make m||n

Step-by-step explanation:

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2 years ago

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klasskru [66]

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7 0

3 years ago

How do you simplify this

Dafna11 [192]

Top:

x / (x + 1) - 1 / x

= [x^2 - (x +1)] / x(x+1)

= (x^2 - x - 1 ) / x (x+1)

Bottom:

x / (x + 1) + 1 / x

= [x^2 + (x +1)] / x(x+1)

= (x^2 + x + 1 ) / x (x+1)

Now you have:

(x^2 - x - 1 ) / x (x+1)

----------------------------

(x^2 + x + 1 ) / x (x+1)

= (x^2 - x - 1 ) / x (x+1) * x (x+1) / (x^2 + x + 1 )

= (x^2 - x - 1 ) /(x^2 + x + 1 )

Answer:

x^2 - x - 1

---------------------

x^2 + x + 1

7 0

3 years ago

Calculate the area of triangle ABC with altitude CD, given A(−7, −1), B(−1, 5), C(0, 0), and D(−3, 3)

Snezhnost [94]

Answer:

A=25 (Rounded)

Step-by-step explanation:

The first step is to find the length of BA which is the base, then you have the find the length of CD, the hight.

To find the distance, you need to use the distance formula which is

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ (It does not matter if A is x1 or is B is x1, it just has to remain consistent)

I am going to first find BA

$d=\sqrt{\left({-7}-{-1}\right)^{2}+\left({-1}-{5}\right)^{2}}$

Next, subtract accordingly

$d=\sqrt{\left({-7}+{1}\right)^{2}+\left({-1}-{5}\right)^{2}}$

$d=\sqrt{\left({-6}\right)^{2}+\left({-6}\right)^{2}}$

Once you do that, you want to square -6 and -6 which will give you

$d=\sqrt{\left({36}\right)+\left({36}\right)}$

Now add 36 with 36 which will leave you with

$d=\sqrt{{72}}$

Do that same thing for CD. If you did it right, you should be left with

$d=\sqrt{{18}}$

to find the area of a triangle, you must use the formula

$A=\frac{B*H}{2}$

The Base is $\sqrt{{72}}$ (BA) and the Hight is $\sqrt{{18}}$ (CD). substitute to get

$A=\frac{\sqrt{{72}}*\sqrt{{18}}}{2}$

Multiply the Base by the Hight to get

$A=\frac{\sqrt{{1,296}}}{2}$

Divide 1,296 by 2 to get

$A=\sqrt{{648}}$

Which is equal to

$A=25.455844122715710878430397035775$

Rounding it to the nearest whole number you get

A=25

Hope this helps. Feel free to ask any follow-up questions.

Have a good day! :)

6 0

3 years ago

A student walk 60m on a bearing

Katena32 [7]

Answer:

d = 234.6 m

Step-by-step explanation:

You can consider a system of coordinates with its origin at the beginning of the walk of the student.

When she start to walk, she is at (0,0)m. After her first walk, her coordinates are calculated by using the information about the incline and the distance that she traveled:

$x_1=60cos28\°=52.97m\\\\y_1=60sin28\°=28.16m$

she is at the coordinates (52.97 , 28.16)m.

Next, when she walks 180m to the east, her coordinates are:

(52.97+180 , 28.16)m = (232.97 , 28.16)m

To calculate the distance from the final point of the student to the starting point you use the Pythagoras generalization for the distance between two points:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}\\\\x=232.97\\\\x_o=0\\\\y=28.16\\\\y_o=0\\\\d=\sqrt{(232.97-0)^2+(28.16-0)^2}m=234.6m$

The displacement of the student on her complete trajectory was of 234.6m

8 0

3 years ago