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Firdavs [7]
3 years ago
14

1200((1)/(2))^(50)/(14)

Mathematics
2 answers:
grigory [225]3 years ago
8 0

Answer:

7.61

Step-by-step explanation:

Airida [17]3 years ago
3 0

Answer:

aprox. 100.94

Step-by-step explanation:

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If the functions y=9/2x^2 Was placed in the form y=ax^b , where a and b are real numbers, then which of the following is the val
exis [7]

Answer:

\frac{13}{2}

Step-by-step explanation:

We have that:

a=\frac{9}{2} \\b=2

Now we can find a+b

\frac{9}{2}+2=\frac{9+4}{2}=\frac{13}{2}

5 0
3 years ago
The total cost of n shirts is $15. The shirts are priced at a constant rate of $3 each.
Flauer [41]

Answer:

3*x=15

Step-by-step explanation:

i think-

5 0
3 years ago
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What is the missing angle
igomit [66]
The missing angle is 53 degrees (82+45=127) (180-127=53)
7 0
3 years ago
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A^2-a+12=0 I need the roots
kumpel [21]

Answer:

a = 1/2 (1 ±sqrt(47))

Step-by-step explanation:

a^2-a+12=0

We will complete the square

Subtract 12 from each side

a^2-a+12-12=0-12

a^2-a=-12

The coefficient of a = -1

-Divide by 2 and then square it

(-1/2) ^2 = 1/4

Add it to each side

a^2 -a +1/4=-12 +1/4

(a-1/2)^2 = -11 3/4

(a-1/2)^2= -47/4

Take the square root of each side

sqrt((a-1/2)^2) =sqrt(-47/4)

a-1/2 = ±i sqrt(1/4) sqrt(47)

a-1/2= ±i/2 sqrt(47)

Add 1/2 to each side

a-1/2+1/2 = 1/2± i/2 sqrt(47)

a =  1/2± i/2 sqrt(47)

a = 1/2 (1 ±sqrt(47))

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
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