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Black_prince [1.1K]
2 years ago
14

Subtract: 4x/x^2-25 - 2/x^2+2x-15

Mathematics
1 answer:
Alex777 [14]2 years ago
4 0

Answer: \dfrac{4x^2-14x+10}{(x-5)(x+5)(x-3)}

Step-by-step explanation:

Given

Subtract the expression

\Rightarrow \dfrac{4x}{x^2-25}-\dfrac{2}{x^2+2x-15}\\\\\Rightarrow \dfrac{4x}{(x-5)(x+5)}-\dfrac{2}{x^2+5x-3x-15}\\\\\Rightarrow \dfrac{4x}{(x-5)(x+5)}-\dfrac{2}{(x+5)(x-3)}\\\\\text{Take the LCM and subtract}\\\\\Rightarrow \dfrac{4x(x-3)-2(x-5)}{(x-5)(x+5)(x-3)}\\\\\Rightarrow \dfrac{4x^2-12x-2x+10}{(x-5)(x+5)(x-3)}\\\\\Rightarrow \dfrac{4x^2-14x+10}{(x-5)(x+5)(x-3)}

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How do I solve this equation
Anestetic [448]
-x - y = 8
2x - y = -1

Ok, we are going to solve this in 2 parts.  First we have to solve for one of the variables in one of the equation in terms of the other variable.  I like to take the easiest equation first and try to avoid fractions, so let's use the first equation and solve for x.

-x - y = 8      add y to each side
-x = 8 + y      divide by -1
x = -8 - y

So now we have a value for x in terms of y that we can use to substitute into the other equation.  In the other equation we are going to put -8 - y in place of the x.

2x - y = -1
2(-8 - y) - y = -1      multiply the 2 through the parentheses
-16 - 2y - y = -1      combine like terms
-16 - 3y = -1            add 16 to both sides
-3y = 15                   divide each side by -3
y = -5

Now we have a value for y.  We need to plug it into either of the original equations then solve for x.  I usually choose the most simple equation.

-x - y = 8
-x - (-5) = 8            multiply -1 through the parentheses
-x + 5 = 8                subtract 5 from each side
-x = 3                      divide each side by -1
x = -3

So our solution set is

(-3, -5)

That is the point on the grid where the 2 equations are equal, so that is the place where they intersect.

4 0
3 years ago
What is the measure of arc CD?
CaHeK987 [17]

Option C:

The measure of arc CD is 40°.

Solution:

Given data:

m∠X = 11° and m(arc AB) = 18°

To find the measure of arc CD:

We know that,

<em>Angle formed by two intersecting secants outside the circle is equal to half of the difference between the intercepted arcs.</em>

$m \angle X =\frac{1}{2}({arc  \ CD}-{arc}  \ AB)

$ 11^\circ =\frac{1}{2}({arc}  \ CD-{18^\circ})

Multiply by 2 on both sides.

22° = arc CD - 18°

Add 18° from both sides.

40° = arc CD

Switch the sides.

arc CD = 40°

Hence the measure of arc CD is 40°.

Option B is the correct answer.

7 0
3 years ago
One of the events in the Winter Olympics is the Men's 500-meter Speed Skating.
poizon [28]

Answer:

Mean: \bar x = 40.61

Median =40.2

Mode = 43.4

Step-by-step explanation:

Given

See attachment for data

Solving (a): The mean

Mean is calculated as:

\bar x = \frac{\sum x}{n}

From the attached:

n = 14

So, the mean is:

\bar x = \frac{43.4+43.4+43.4+43.1+43.2+40.2+40.2+40.1+40.3+39.44+39.17+38.03+38.19+36.45}{14}

\bar x = \frac{568.58}{14}

\bar x = 40.61

Solving (b): The median

n = 14; this is an even number. So, the median is:

Median = \frac{1}{2}(n+1)

Median = \frac{1}{2}(14+1)

Median = \frac{1}{2}(15)

Median = 7.5th

This implies that the median is the average of the 7th and 8th item.

Next, is to order the data (in ascending order): <em>36.45, 38.03, 38.19, 39.17, 39.44, 40.1, 40.2, 40.2, 40.3, 43.1, 43.2, 43.4, 43.4, 43.4.</em>

The 7th and 8th items are: 40.2 and 40.2

The median is:

Median = \frac{1}{2}(40.2 + 40.2)

Median = \frac{1}{2}*80.4

Median =40.2

Solving (c): The mode

43.4 has the highest number of occurrence.

So:

Mode = 43.4

8 0
3 years ago
Name
Anvisha [2.4K]

Answer:

4x+8

6x-8

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Step-by-step explanation:

a) 4(x + 2)=4x+8

b) 2(3x - 4)  =6x-8

c) 4(3 - 6x) 2x + 18-2(2 – 3x)  

=(12-24x)2x-4+6x

=24x-48x^2-4+6x

=-48x^2+30x-4

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3 years ago
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irakobra [83]
We’re told that QR and RS are equal, so this is an isosceles triangle.

Meaning the two base angles must also be equal

Using this information, we know that the third angle must be 180 - (22 + 22), since the angles in a triangle add up to 180

8 0
2 years ago
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