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Vika [28.1K]
4 years ago
5

Name the intersection of plane N and line AE

Mathematics
1 answer:
irina1246 [14]4 years ago
3 0
<span>Name the intersection of plane N and line AE is point B.
In General, the intersection of straight line and plane may be:1) one point (as in our case)2) an Infinite number of points - the whole straight line (when the straight line belongs to the plane)3) the empty set (when the straight line and plane are parallel to each other)</span>
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3 years ago
At one gas station, gas costs $2.75 per gallon. Write an equation that relates the total cost, C, to the number of gallons of ga
melisa1 [442]

Answer:

The equation will be:

  • C = 2.75g

Step-by-step explanation:

Given that at one station, gas costs $2.75 per gallon.

Let 'g' be the number of gallons of gas purchases

Let 'C' be the total cost

Thus, the expression for total cost can be generated by multiplying the price per gallon by the number of gallons.

Thus,

C = 2.75g

Therefore, the equation will be:

  • C = 2.75g
6 0
3 years ago
Plz with steps plzzzzzz
Stella [2.4K]

Answer:  -\frac{\sqrt{2a}}{8a}

=======================================================

Explanation:

The (x-a) in the denominator causes a problem if we tried to simply directly substitute in x = a. This is because we get a division by zero error.

The trick often used for problems like this is to rationalize the numerator as shown in the steps below.

\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x}-\sqrt{x+a})(\sqrt{3a-x}+\sqrt{x+a})}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x})^2-(\sqrt{x+a})^2}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-(x+a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-x-a}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\

\displaystyle \lim_{x\to a} \frac{2a-2x}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(-a+x)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(x-a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\

At this point, the (x-a) in the denominator has been canceled out. We can now plug in x = a to see what happens

\displaystyle L = \lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\L = \frac{-2}{4(\sqrt{3a-a}+\sqrt{a+a})}\\\\\\L = \frac{-2}{4(\sqrt{2a}+\sqrt{2a})}\\\\\\L = \frac{-2}{4(2\sqrt{2a})}\\\\\\L = \frac{-2}{8\sqrt{2a}}\\\\\\L = \frac{-1}{4\sqrt{2a}}\\\\\\L = \frac{-1*\sqrt{2a}}{4\sqrt{2a}*\sqrt{2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{2a*2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{(2a)^2}}\\\\\\L = \frac{-\sqrt{2a}}{4*2a}\\\\\\L = -\frac{\sqrt{2a}}{8a}\\\\\\

There's not much else to say from here since we don't know the value of 'a'. So we can stop here.

Therefore,

\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)} = -\frac{\sqrt{2a}}{8a}\\\\\\

3 0
3 years ago
A certain circle can be represented by the following equation. x^2+y^2+6y-72=0 What is the center of this circle ?
jek_recluse [69]

Answer:

(0, -3)

Step-by-step explanation:

Here we'll rewrite x^2+y^2+6y-72=0 using "completing the square."

Rearranging x^2+y^2+6y-72=0, we get  x^2 + y^2 + 6y                = 72.

x^2 is already a perfect square.  Focus on rewriting y^2 + 6y as the square of a binomial:  y^2 + 6y becomes a perfect square if we add 9 and then subtract 9:

x^2 + y^2 + 6y + 9  - 9              = 72:

x^2 + (y + 3)^2 = 81

Comparing this to the standard equation of a circle with center at (h, k) and radius r,

(x - h)^2 + (y - k)^2 = r^2.  Then h = 0, k = -3 and r = 9.

The center of the circle is (h, k), or (0, -3).

8 0
3 years ago
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