Do y’all know the answer to this!?

A turtle was swimming 2 feet below the surface of a pond. While looking for food, he went

Snezhnost [94]

Answer:

9 feet down

Step-by-step explanation:

2+7=9

3 0

3 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

4 years ago

The length of each side of a square was decreased by 2 inches so the perimeter is now 49 inches. What was the original length of

aleksley [76]

The Original length of each side of square was 14.25 inches.

Step-by-step explanation:

Perimeter of a square is given by:

Perimeter = 4 * side

Let x be the original length of each side of square

So

decreasing 2 inch will mean

$x-2$

Perimeter of square after reducing 2 inches from each side

$4(x-2) = 49\\4x-8 = 49\\4x = 49+8\\4x = 57\\\frac{4x}{4} = \frac{57}{4}\\x = 14.25$

So,

The Original length of each side of square was 14.25 inches.

Keywords: Perimeter, square

Learn more about perimeter at:

#LearnwithBrainly

5 0

3 years ago

Find the volume of the bag and candle.Round to the nearest tenth

user100 [1]

The volume of the bag will be found by multiplying the length times the width times the height. This would be 5.5 x 3 x 8. The volume of the bag is 20.9 cubic inches. The volume of the candle would be found by using the formula for finding volume of a cylinder. V =pi(approximately 3.14) x r^2 x h, where r is the radius( half the diameter) and h is the height. V = 3.14 x 1.25 x 1.25 x 6. The volume of the cylinder would be 29.4 cubic inches.

5 0

3 years ago

_ 5x^20-7x^10+15 in quadratic form

Tpy6a [65]

Answer:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -(1/10 (-7 - sqrt(349)))^(1/10) or x = (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10)

Step-by-step explanation:

Solve for x:

-5 x^20 - 7 x^10 + 15 = 0

Substitute y = x^10:

-5 y^2 - 7 y + 15 = 0

Divide both sides by -5:

y^2 + (7 y)/5 - 3 = 0

Add 3 to both sides:

y^2 + (7 y)/5 = 3

Add 49/100 to both sides:

y^2 + (7 y)/5 + 49/100 = 349/100

Write the left hand side as a square:

(y + 7/10)^2 = 349/100

Take the square root of both sides:

y + 7/10 = sqrt(349)/10 or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

y = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Substitute back for y = x^10:

x^10 = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Taking 10^th roots gives (sqrt(349)/10 - 7/10)^(1/10) times the 10^th roots of unity:

x = -(1/10 (sqrt(349) - 7))^(1/10) or x = (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or y = -7/10 - sqrt(349)/10

Substitute back for y = x^10:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x^10 = -7/10 - sqrt(349)/10

Taking 10^th roots gives (-7/10 - sqrt(349)/10)^(1/10) times the 10^th roots of unity:

Answer: x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -(1/10 (-7 - sqrt(349)))^(1/10) or x = (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10)

8 0

3 years ago