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Allisa [31]
3 years ago
10

PLEASE HELP!! 48 POINTS!!

Mathematics
1 answer:
Likurg_2 [28]3 years ago
4 0
The exact number is 121.966019, so if you round to the nearest whole number, the answer is 122.
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Which is the simplified form of the expression?
Scorpion4ik [409]

Answer:

1 / q^30.

Step-by-step explanation:

[(p^2)(q^5)]^-4 * [(p^-4)(q^5)]^-2

Using the law (a^b)^c =  a^bc :-

=  p^-8 * q^-20 * p^8 * q^-10

= p^(-8+8) * q^(-20-10)

= p^0 * q^-30

= 1 * q^-30.

= 1 / q^30.

8 0
3 years ago
Convert the time. Enter your answer in the box.<br><br> 4 minutes 53 seconds = <br> seconds
ch4aika [34]

Answer:

293 seconds

Step-by-step explanation:

60*4=240

240+53=293

3 0
2 years ago
What is area of each circle below?
KiRa [710]

Answer:

Step-by-step explanation: A= (Pi) x r^2

4 0
3 years ago
Given the midpoint (1.5,1.5) and the endpoint (5,7) where is the other endpoint located
earnstyle [38]

The formula of a midpoint:

M_{AB}\left(\dfrac{x_A+x_B}{2},\ \dfrac{y_A+y_B}{2}\right)

We have:

M(1.5,\ 1.5)\to x_M=1.5,\ y_M=1.5\\A(5,\ 7)\to x_A=5,\ y_A=7

Substitute

\dfrac{5+x_B}{2}=1.5\qquad|\cdot2\\\\5+x_B=3\qquad|-5\\\\x_B=-2\\\\\dfrac{7+y_B}{2}=1.5\qquad|\cdot2\\\\7+y_B=3\qquad|-7\\\\y_B=-4

<h3>Answer: (-2, -4)</h3>
8 0
3 years ago
If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form
algol13

Answer:

\frac{d^2y}{dx^2} = \frac{-4}{3}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Product Rule: \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Chain Rule: \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Quotient Rule: \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

  1. Implicit Differentiation [Basic Power Rule]:                                                  -y'-6x^2=2yy'
  2. [Algebra] Isolate <em>y'</em> terms:                                                                              -6x^2=2yy'+y'
  3. [Algebra] Factor <em>y'</em>:                                                                                       -6x^2=y'(2y+1)
  4. [Algebra] Isolate <em>y'</em>:                                                                                         \frac{-6x^2}{(2y+1)}=y'
  5. [Algebra] Rewrite:                                                                                           y' = \frac{-6x^2}{(2y+1)}

<u>Step 3: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

  1. Differentiate [Quotient Rule/Basic Power Rule]:                                          y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}
  2. [Derivative] Simplify:                                                                                       y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}
  3. [Derivative] Back-Substitute <em>y'</em>:                                                                     y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}
  4. [Derivative] Simplify:                                                                                      y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}

<u>Step 4: Find Slope at Given Point</u>

  1. [Algebra] Substitute in <em>x</em> and <em>y</em>:                                                                     y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}
  2. [Pre-Algebra] Exponents:                                                                                      y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}
  3. [Pre-Algebra] Multiply:                                                                                   y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}
  4. [Pre-Algebra] Add:                                                                                         y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}
  5. [Pre-Algebra] Exponents:                                                                               y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}
  6. [Pre-Algebra] Divide:                                                                                      y''(-1,-2) = \frac{-36+24 }{9}
  7. [Pre-Algebra] Add:                                                                                          y''(-1,-2) = \frac{-12}{9}
  8. [Pre-Algebra] Simplify:                                                                                    y''(-1,-2) = \frac{-4}{3}
6 0
2 years ago
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