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Nastasia [14]
3 years ago
14

BRAINLIEST I forgot how to tell time so maybe show work?

Mathematics
2 answers:
4vir4ik [10]3 years ago
7 0

Answer:

1 hour 37 mins

Step-by-step explanation:

yKpoI14uk [10]3 years ago
5 0

Answer:

1 hour and 37 minutes

Step-by-step explanation:

hope u have a great day! you are so smart! :)

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6 0
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What the percentage of 25 out of 100?
Aleks04 [339]

Answer:

25%

Step-by-step explanation:

Convert fraction (ratio) 25 / 100 Answer: 25%

8 0
3 years ago
Read 2 more answers
Brad is putting up a 10 foot diameter circular flower garden in his yard. He will put plastic edging along the flowerbed. How ma
Ymorist [56]
The answer is circled in red. Hope this helps!

4 0
3 years ago
Use the limit definition of the derivative to find the slope of the tangent line to the curve
ale4655 [162]

Answer:

\displaystyle f'(4) = 63

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> </u>

Distributive Property

<u>Algebra I</u>

  • Expand by FOIL (First Outside Inside Last)
  • Factoring
  • Function Notation
  • Terms/Coefficients

<u>Calculus</u>

Derivatives

The definition of a derivative is the slope of the tangent line.

Limit Definition of a Derivative: \displaystyle f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}  

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = 7x² + 7x + 3

Slope of tangent line at x = 4

<u>Step 2: Differentiate</u>

  1. Substitute in function [Limit Definition of a Derivative]:                              \displaystyle f'(x)= \lim_{h \to 0} \frac{[7(x + h)^2 + 7(x + h) + 3]-(7x^2 + 7x + 3)}{h}
  2. [Limit - Fraction] Expand [FOIL]:                                                                    \displaystyle f'(x)= \lim_{h \to 0} \frac{[7(x^2 + 2xh + h^2) + 7(x + h) + 3]-(7x^2 + 7x + 3)}{h}
  3. [Limit - Fraction] Distribute:                                                                            \displaystyle f'(x)= \lim_{h \to 0} \frac{[7x^2 + 14xh + 7h^2 + 7x + 7h + 3] - 7x^2 - 7x - 3}{h}
  4. [Limit - Fraction] Combine like terms (x²):                                                     \displaystyle f'(x)= \lim_{h \to 0} \frac{14xh + 7h^2 + 7x + 7h + 3 - 7x - 3}{h}
  5. [Limit - Fraction] Combine like terms (x):                                                      \displaystyle f'(x)= \lim_{h \to 0} \frac{14xh + 7h^2 + 7h + 3 - 3}{h}
  6. [Limit - Fraction] Combine like terms:                                                           \displaystyle f'(x)= \lim_{h \to 0} \frac{14xh + 7h^2 + 7h}{h}
  7. [Limit - Fraction] Factor:                                                                                 \displaystyle f'(x)= \lim_{h \to 0} \frac{h(14x + 7h + 7)}{h}
  8. [Limit - Fraction] Simplify:                                                                               \displaystyle f'(x)= \lim_{h \to 0} 14x + 7h + 7
  9. [Limit] Evaluate:                                                                                                 \displaystyle f'(x) = 14x + 7

<u>Step 3: Find Slope</u>

  1. Substitute in <em>x</em>:                                                                                                \displaystyle f'(4) = 14(4) + 7
  2. Multiply:                                                                                                           \displaystyle f'(4) = 56 + 7
  3. Add:                                                                                                                  \displaystyle f'(4) = 63

This means that the slope of the tangent line at x = 4 is equal to 63.

Hope this helps!

Topic: Calculus AB/1

Unit: Chapter 2 - Definition of a Derivative

(College Calculus 10e)

3 0
3 years ago
Consider the expansion of (5p + 2q)^6. Determine the coefficients for the terms with the powers of p and q shown.
Step2247 [10]

Answer:

Remember, the expansion of (x+y)^n is (x+y)^n=\sum_{k=0}^n \binom{n}{k}x^{n-k}y^k, where \binom{n}{k}=\frac{n!}{(n-k)!k!}.

Then,

(5p+2q)^6=\sum_{k=0}^6\binom{6}{k}(5p)^{6-k}(2q)^k=\sum_{k=0}^6\binom{6}{k}5^{6-k}2^k p^{6-k}q^k

Then, the coefficient of the term p^{6-k}q^k is \binom{6}{k}5^{6-k}2^k

a) since 6-k=2, then k=4. So the coefficient of p^2q^4 is

\binom{6}{4}5^{6-4}2^4=15*5^2*2^4=15*25*16=6000

b) since 6-k=5, then k=1. So, the coefficient of p^5q is

\binom{6}{1}5^{6-1}2^1=6*5^5*2=37500

c) since 6-k=3, then k=3. So, the coefficient of p^3q^3 is

\binom{6}{3}5^{6-3}2^3=20*5^3*8=20000

6 0
3 years ago
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