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Len [333]
3 years ago
6

PLEASE HELP ME WITH IT.

Mathematics
1 answer:
scoundrel [369]3 years ago
7 0

OK.  I did it.  Now let's see if I can go through it without
getting too complicated.

I think the key to the whole thing is this fact:

     A radius drawn perpendicular to a chord bisects the chord.

That tells us several things:

-- OM bisects AB. 
   'M' is the midpoint of AB.
   AM is half of AB.

-- ON bisects AC.
    'N' is the midpoint of AC.
   AN is half of AC.

--  Since AC is half of AB,
     AN is half of AM.
     a = b/2 

Now look at the right triangle inside the rectangle.
'r' is the hypotenuse, so

                                            a² + b² = r²

But  a = b/2, so             (b/2)² + b² = r²

(b/2)² = b²/4                   b²/4   + b² = r²

Multiply each side by 4:     b² + 4b² = 4r²
                                       -  -  -  -  -  -  -  -  -  -  -
                                            0  + 5b² = 4r²  
Repeat the
original equation:                a² +  b² =  r²

Subtract the last
two equations:                  -a² + 4b² = 3r² 

Add  a²  to each side:              4b²  =  a² + 3r² .    <=== ! ! !
 
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Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant
Alex787 [66]

Find the intercepts for both planes.

Plane 1, <em>x</em> + <em>y</em> + 2<em>z</em> = 2:

y=z=0\implies x=2\implies (2,0,0)

x=z=0\implies y=2\implies(0,2,0)

x=y=0\implies 2z=2\implies z=1\implies(0,0,1)

Plane 2, 4<em>x</em> + 4<em>y</em> + <em>z</em> = 8:

y=z=0\implies4x=8\implies x=2\implies(2,0,0)

x=z=0\implies4y=8\impliesy=2\implies(0,2,0)

x=y=0\implies z=8\implies(0,0,8)

Both planes share the same <em>x</em>- and <em>y</em>-intercepts, but the second plane's <em>z</em>-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (<em>x</em>, <em>y</em>)-plane where <em>z</em> = 0, we see the bounded region projects down to the triangle in the first quadrant with legs <em>x</em> = 0, <em>y</em> = 0, and <em>x</em> + <em>y</em> = 2, or <em>y</em> = 2 - <em>x</em>.

So the volume of the region is

\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx

=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx

=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}

3 0
3 years ago
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kolezko [41]

Answer:

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4 0
3 years ago
Read 2 more answers
Write g(x)=4x^2+88 in vertex form
umka21 [38]

Answer:

y

=

4

(

x

+

11

)

2

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484

Step-by-step explanation:

5 0
4 years ago
In a game called Taxation and Evasion, a player rolls a pair of dice. If, on any turn, the sum is 7, 11, or 12, the player gets
AnnZ [28]

Answer:

Probability = \frac{243}{1024}

Step-by-step explanation:

Given

Sum = \{7, 11, 12\}

Rolls = 5

Required

Probability of not getting audited

If a pair of dice is rolled, the following are the observations of the sum

Outcomes = 36

Sum\ of\ 7 = 6

Sum\ of\ 11 = 2

Sum\ of\ 12 = 1

So, in a single roll; The probability of getting audited is:

p= \frac{1 + 2 + 6}{36}

p= \frac{9}{36}

p= \frac{1}{4}

The probability of not getting audited in a single roll is:

q = 1 - p --- Complement rule

q = 1 - \frac{1}{4}

Take LCM

q = \frac{4 - 1}{4}

q = \frac{3}{4}

The probability of not getting audited in 5 rolls is:

Probability = q^5

Probability = (\frac{3}{4})^5

Probability = \frac{3^5}{4^5}

Probability = \frac{243}{1024}

7 0
3 years ago
Sally’s dance school had 60 students last year. This year there are only 48 students enrolled. By what percent did the enrollmen
Morgarella [4.7K]

Answer:

-20%

My brain is weird on how i figure it out but I divided 48 by 60 and got .80 so i just got the other whole to make it 1 so it is 20%. This is not the correct way to do this but this is how i got my answer.

8 0
3 years ago
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