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3 years ago

Please help on this one

Mathematics

2 answers:

kondor19780726 [428]3 years ago

6 0

The answer is D because it passes the horizontal line test.

Artemon [7]3 years ago

3 0

B because a horizontal line

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Can anyone please help me with this I have to get it done before tomorrow

lions [1.4K]

Titania is the greatest because it has the largest magnitude(exponent). than obreron than ariel than umbriel and finally miranda

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3 years ago

Question 8 (1 point)

sergey [27]

Y=-5x+4 y=7x+5 that is the line

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:

Calculus

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:

Step 1: Define

Identify

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

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

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

Step 3: Integrate Pt. 2

[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

2 years ago

3(2−4)=−30 step by step

Margarita [4]

3•2-3•4
6-12=-30
-6=-30
false

6 0

3 years ago

A parking lot charges $3 to park a car for the first hour and $2 per hour after that. If you use more than one parking space, th

fenix001 [56]

Consider the charge for parking one car for t hours.

If t is more than 1, then the function is y=3+2(t-1), because 3 $ are payed for the first hour, then for t-1 of the left hours, we pay 2 $.

If t is one, then the rule y=3+2(t-1) still calculates the charge of 3 $, because substituting t with one in the formula yields 3.

75% is 75/100 or 0.75.

For whatever number of hours t, the charge for the first car is 3+2(t-1) $, and whatever that expression is, the price for the second car and third car will be

0.75 times 3+2(t-1). Thus, the charge for the 3 cars is given by:

3+2(t-1)+0.75[3+2(t-1)]+0.75[3+2(t-1)]=3+2(t-1)+0.75 × 2[3 + 2(t − 1)].

Thus, the function which total parking charge of parking 3 cars for t hours is:

f(t) = (3 + 2(t − 1)) + 0.75 × 2(3 + 2(t − 1))

Answer: C

4 0

3 years ago