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Elenna [48]
3 years ago
8

D-c/pie solve for pie

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
6 0
With a little algebraic manipulation we can solve for pi...

d-c / pi

we want to get pi by itself so first we multiply both sides by pi to get it out of the denominator...

pi × (d-c) = pi / pi
pi × (d-c) = 1

Now get pi by itself by dividing both sides by d-c...

pi × (d-c) / d-c = 1 / d-c

simplify

pi × 1 = 1 / d-c
pi = 1 / d-c

Done!
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A poster is to have a total area of 125 cm2. There is a margin round the edges of 6 cm at the top and 4 cm at the sides and bott
jeka57 [31]

Answer:

The correct answer is 15 cm.

Step-by-step explanation:

Let the width of the required poster be a cm.

We need to have a 6 cm margin at the top and a 4 cm margin at the bottom. Thus total margin combining top and bottom is 10 cm.

Similarly total margin combining both the sides is (4+4=) 8 cm.

So the required printing area of the poster is given by {( a-10 ) × ( a - 8) } cm^{2}

This area is equal to 125 cm^{2} as per as the given problem.

∴ (a - 10) × (a - 8) = 125

⇒ a^{2} - 18 a +80 -125 =0

⇒ a^{2} - 18 a -45 = 0

⇒ (a-15) (a-3) = 0

By law of trichotomy the possible values of a are 15 and 3.

But a=3 is absurd as a > 4.

Thus the required answer is 15 cm.

7 0
2 years ago
Read 2 more answers
This is Matrix for pre calc
gavmur [86]

The given system of equations in augmented matrix form is

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]

If you need to solve this, first get the matrix in RREF:

  • Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]

  • Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]

  • Add 164(row 3) to -91(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]

  • Multiply row 4 by 1/13080:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]

  • Add -153(row 4) to row 3:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]

  • Multiply row 3 by -1/91:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add 6(row 3) and -8(row 4) to row 2:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 2 by 1/5:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 1 by 1/3:

\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

So the solution to this system is (w,x,y,z)=(1,-2,4,-3).

6 0
3 years ago
The point (2, 3) is on the line given by which equation below?
love history [14]
Plug in the points and see if y and x are equal to each other.

(2,3)
y= x-1
3= 2-1 
3= 1 (not the solution)

y=3x
3=3*3
3=9 (not the solution)

y=x+1
3=2+1
3=3 (solution) We can check to make sure.

y=-3x
3=-3*3
3=-9 (not the solution)

y=x+1 <em>is the answer.
</em>

I hope this helps!
~kaikers


5 0
3 years ago
Read 2 more answers
The median of the data set is 81. What number is missing?<br><br> 82, 87, _, 77, 70, 80, 88
mina [271]
Median is the middle number.
Rearrange the numbers to smallest to biggest.
70, 77, 80, _, 82, 87, 88
If the median of the data set is 81 the middle number needs to be 81.
So the missing number is 81.
70, 77, 80, 81, 82, 87, 88
7 0
3 years ago
Read 2 more answers
HELP HELP HELP!!!!!!!
erma4kov [3.2K]

Answer:

b = 77

Step-by-step explanation:

a² + b² = c²

36² + b² = 85²

1296 + b² = 7225

b² = 7225 - 1296

b² = 5929

\sqrt{b^2} =\sqrt{5929}

b = 77

3 0
2 years ago
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