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malfutka [58]
3 years ago
5

A reflecting telescope contains a mirror shaped like a parabola of revolution. If the mirror is 16 inches across at its opening

and is 2 feet deep, where will the light be concentrated?
Mathematics
1 answer:
choli [55]3 years ago
5 0

Answer:

18

Step-by-step explanation:

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Please help asap calculate the slope of the line through the points (6,-8) and (3,-4).
Taya2010 [7]

Answer:

Step-by-step explanation:

Mathematically, we can calculate the slope

using the formula below;

m = (y2-y1)/(x2-x1)

Where (x1.y1) = (6,-8)

(x2, y2) = (3,-4)

So inputing these values into the formula, we have

m = (-4-(-8))/(3-6)

m = (-4+8)/-3 = 4/-3 = -4/3

Now, we want to write the equation of the line;

y= -4/3x + c

to get the value of c, we simply need to use any of the points. We use any of the two points by substituting the values of x and y at that point.

Let’s use (3,-4)

That would give ;

-4 = -4/3(3) + c

-4 = -4 + c

This means c = 0

Thus the equation of the line is y = -4/3x

Now to find out if the point (-3,4) is on the line

We only need to substitute the values of x and y here and see ;

Thus;

4 = -4/3(-3)

Sine 4 = 4

We can conclude that the point is on the line in question

7 0
3 years ago
Cynthia was asked to subtract 0.223 from 0.54. Which of the following is the best estimate that cynthia can make to check her an
Zigmanuir [339]
I need help with this one too , can you guys please help us


In desperate need

Answer :
7 0
3 years ago
I need to know the answer to this lol
asambeis [7]
Length of GH is also 11 meters
5 0
3 years ago
Read 2 more answers
What is the LEAST amount of information you need about a pair of triangles to be SURE they are congruent?
tigry1 [53]
I hope you have heard about the congruency rules like SSS, SAS, ASA

Those are the least amount of information you need
3 0
3 years ago
Read 2 more answers
Find the following integral
ololo11 [35]

There's nothing preventing us from computing one integral at a time:

\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2

\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2

\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx

Expand the integrand completely:

x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x

Then

\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}

4 0
3 years ago
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