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melamori03 [73]
4 years ago
8

If x ≠ -2, then 5x2 -20/5x+10 =​

Mathematics
1 answer:
MissTica4 years ago
6 0

Answer:

x-2\, ,x\neq -2

Step-by-step explanation:

So we have the expression:

\frac{5x^2-20}{5x+10}\, ,x\neq -2

For both the numerator and the denominator, factor out a 5:

\frac{5(x^2-4)}{5(x+2)}\, ,x\neq-2

The term in the numerator can be factored. This is the difference of two squares pattern:

\frac{5(x-2)(x+2)}{5(x+2)}\, ,x\neq-2

Both layers have a 5(x+2). Cancel them, Since we know that x <em>cannot</em> be -2, we can safely do so:

x-2\, ,x\neq -2

And we're done!

Notes:

If we weren't told that x ≠ -2, then we cannot divide them because if x <em>did</em> equal -2, the equation would be undefined. However, since we were given that, we can simplify.

However, even if we weren't given x ≠ -2, we can <em>still</em> simplify, but we need to add the restraint ourselves. This must be done, or else the equation won't be correct.

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Solve the following pair of equations by elimination method: x+y=4, 2x+y=3​
allsm [11]

Answer:

x = 1

y = 5

Step-by-step explanation:

x + y = 4 __ (1)

2x + y = 3 __ (2)

equation (1) x 2, (2) x 1

2x + 2y = 8

2x + y = 3

0  + y = 5

y= 5 (ANS)

x + y = 4

x + 5 = 4

x = 5 - 4

x = 1 (ANS)

I HOPE MY ANSWER IS CORRECT IF NOT I APOLOGIZE.

5 0
2 years ago
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olganol [36]

Answer:

What is

3

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8 0
3 years ago
Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span><1,2,−2>⋅<x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
<span>→<span>P1</span><span>(0,0,−<span>12</span>)</span></span>

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

<span><span>D=<span><span>|a<span>x1</span>+b<span>y1</span>+c<span>z1</span>+d|</span><span>√<span><span>a2</span>+<span>b2</span>+<span>c2</span></span></span></span></span>
<span>2=<span><span><span>∣∣</span>1⋅0+2⋅0+<span>(−2)</span>⋅<span>(−<span>12</span>)</span>+d<span>∣∣</span></span><span>√<span>1+4+4</span></span></span></span>
<span><span>|d+1|</span>=2⋅3</span> => <span><span>|d+1|</span>=6</span>First solution:
<span>d+1=6</span> => <span>d=5</span>
<span>→x+2y−2z+5=0</span>Second solution:
<span>d+1=−6</span> => <span>d=−7</span>
<span>→x+2y−2z−7=<span>0</span></span></span>
8 0
3 years ago

larisa86 [58]

Answer:

The correct answer is D, 0.20. You add the probabilities of either situation occurring and subtract the probability of both of them occurring.

Step-by-step explanation:

0.25 + 0.08 - 0.13 = 0.20

3 0
3 years ago
Read 2 more answers
What is the value of the expression<br> when x = 12 and p = -4?
goblinko [34]

Answer:

-48

Step-by-step explanation:

4 0
3 years ago
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