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3 years ago

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True or False: On the Mertler’s vacation to Florida, they drove 180 miles in 3 hours before stopping for lunch. After lunch they

drove 120 miles in 2 hours before stopping for gas. Are these rates equivalent? Explain your reasoning.

1 answer:

natita [175]3 years ago

7 0

Answer:

true

Step-by-step explanation:

because the ratios are both 60 miles in 1 hour

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60/1 in a equivalent ratio

Elena-2011 [213]

60 /1 = 120 /2 = 180 /3

hope it helps

8 0

3 years ago

Read 2 more answers

Triangle LMN has L at (1, -1) and M at (2, 3). Triangle L'M'N' has L' at (-1, -1), M' at (3, -2), and N' is at (-3, 0). What are

fgiga [73]

Answer:

The answer is N(0,-3)

3 0

3 years ago

EF is the median of trapezoid ABCD.

BARSIC [14]

Answer:

Part A) $x = 11$

B) length of BC = x + 12 = x + 12 = 11 + 12 =23

length of EF = 4x - 18 = 4(11) - 18 = 44 - 18 =26

length of AD = 3x - 18 = 3(11) - 4 = 33 - 4 =29

Step-by-step explanation:

Median of trapezium is $m=\frac{base1+base2}{2}$

In provided figure, base1 is x+12 and base2 is 3x-4

A) solve for value of x

calculate the median $m=\frac{base1+base2}{2}$

$4x-18=\frac{x+12+3x-4}{2}$

$4x-18=\frac{4x+8}{2}$

$4x-18=2x+4$

$2x-18=4$

$2x = 18 + 4$

$2x = 22$

$x = 11$

B) To find the length of BC, AD and EF , put the value of x in equation of lines

length of BC = x + 12 = x + 12 = 11 + 12 =23

length of EF = 4x - 18 = 4(11) - 18 = 44 - 18 =26

length of AD = 3x - 4 = 3(11) - 4 = 33 - 4 =29

5 0

4 years ago

What is the equation of the line?

Neporo4naja [7]

Y=4 because y is 4 all the time

3 0

3 years ago

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

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

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 4x^{-2}$
[<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

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

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

3 years ago