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aev [14]
3 years ago
8

Passing through (12,3) perpendicular to y + 3x = 4

Mathematics
1 answer:
enot [183]3 years ago
3 0
Remember the formula?
m1m2=-1

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An insured's roof cost $4,000 when installed 5 years ago. It has been damaged by hail and must be replaced. The new roof will co
hoa [83]

Answer:

ACV=$4,500

Step-by-step explanation:

We have that the actual cash value (ACV) is defined as:

ACV=\dfrac{R\times(E-C)}{E}

Where:

ACV = actual cash value

R = replacement cost or purchase price of the item

E = expected life of the item

C = current life of the item

Then we have R=$6,000, C=5years, and to find the expected life of the item we can use the depreciating of the roof, then if the roof is depreciating $200 each year we just need to divide $4,000 by $200 to find the expected life of the roof:

\dfrac{4,000}{200}=20

Then the espected life of the roof is 20 years, with this result we have all the data, then:

ACV=\dfrac{\$6,000\times (20-5)}{20}=\dfrac{\$6,000\times (15)}{20}=\dfrac{\$90,000}{20}=\$4,500

Then the ACV is $4,500

5 0
3 years ago
Conditional probabilities are based on some event occurring given that something else has already occurred?
alex41 [277]

The answer is true. A conditional probability is a measure of the probability of an event given that (by assumption, presumption, assertion or evidence) another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A in the condition B", is usually written as P (A|B). The conditional probability of A given B is well-defined as the quotient of the probability of the joint of events A and B, and the probability of B.

8 0
3 years ago
HELP AP Math
Afina-wow [57]
4 to the tenth power
3 0
3 years ago
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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:

  1. f(x) = cxⁿ
  2. 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>

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 4x^{-2}
  2. [<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]:                       \displaystyle du = \frac{-8}{x^3} \ dx
  3. [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>

  1. [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
  2. [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
  3. [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}}
  4. 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)
  5. 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
Am I right or if not what is it. :)
dangina [55]

Answer:

Step-by-step explanation:

That is correct.

P=4s=4(12)=48in

A=s^2=12^2=144in^2

8 0
3 years ago
Read 2 more answers
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