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igomit [66]
3 years ago
10

(Geometry Question) For Tangent A, would it be 1.05 or 0.9524?

Mathematics
2 answers:
ohaa [14]3 years ago
5 0

Answer :

tan(A) = 1.05

Explanation in the above picture

Anna71 [15]3 years ago
3 0
1.05 trust me i just took a test with this and i got it right
You might be interested in
When Aria commutes to work, the amount of time it takes her to arrive is normally distributed with a mean of 28 minutes and a st
MrRa [10]

Answer:

Her commute would be between 32 and 35 minutes 33 times.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 28, \sigma = 4.5

Proportion of days in which the commute is between 32 and 35 minutes:

This is the pvalue of Z when X = 35 subtracted by the pvalue of Z when X = 32.

X = 35

Z = \frac{X - \mu}{\sigma}

Z = \frac{35 - 28}{4.5}

Z = 1.555

Z = 1.555 has a pvalue of 0.94.

X = 32

Z = \frac{X - \mu}{\sigma}

Z = \frac{32 - 28}{4.5}

Z = 0.89

Z = 0.89 has a pvalue of 0.8133.

0.94 - 0.8133 = 0.1267

Out of 262 days:

Each day, 0.1267 probability

0.1267*262 = 33

Her commute would be between 32 and 35 minutes 33 times.

6 0
4 years ago
How does the graph of g(x) = 5x – 3 compare to the graph of f(x) = 5*?
Karo-lina-s [1.5K]

Answer:

the graph is shifted down 3 units

Step-by-step explanation:

5x-3

meaning the original graph of 5x going down three units

3 0
3 years ago
Convert 2 liters to gallons.
Iteru [2.4K]
2 liters to gallons is 0.53 Gallons.
5 0
3 years ago
Find the particular solution of the differential equation?<br> /=5^3+9^2, when =1, =8
kipiarov [429]

Answer:

\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}

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  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

<u>Algebra I</u>

  • Exponential Rule [Rewrite]:                                                                           \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

Derivatives

Derivative Notation

Solving Differentials - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]:                                                               \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

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

Integration Property [Addition/Subtraction]:                                                       \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Step-by-step explanation:

*Note:

Ignore the Integration Constant C on the left hand side of the differential equation when integrating.

<u>Step 1: Define</u>

\displaystyle \frac{ds}{dt} = 5t^3 + \frac{9}{t^2}

t = 1

s = 8

<u>Step 2: Integrate</u>

  1. [Derivative] Rewrite [Leibniz's Notation]:                                                     \displaystyle ds = (5t^3 + \frac{9}{t^2})dt
  2. [Equality Property] Integrate both sides:                                                     \displaystyle \int {} \, ds = \int {(5t^3 + \frac{9}{t^2})} \, dt
  3. [Left Integral] Reverse Power Rule:                                                             \displaystyle s = \int {(5t^3 + \frac{9}{t^2})} \, dt
  4. [Right Integral] Rewrite [Integration Property - Addition]:                           \displaystyle s = \int {5t^3} \, dt + \int {\frac{9}{t^2}} \, dt
  5. [Right Integrals] Rewrite [Integration Property - Multiplied Constant]:     \displaystyle s = 5\int {t^3} \, dt + 9\int {\frac{1}{t^2}} \, dt
  6. [Right Integrals] Rewrite [Exponential Rule - Rewrite]:                               \displaystyle s = 5\int {t^3} \, dt + 9\int {t^{-2}} \, dt
  7. [Right Integrals] Reverse Power Rule:                                                         \displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{t^{-1}}{-1}) + C
  8. [Right Integrals] Rewrite [Exponential Rule - Rewrite]:                               \displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{1}{t}) + C
  9. Multiply:                                                                                                         \displaystyle s = \frac{5t^4}{4} + \frac{9}{t} + C

<u>Step 3: Solve</u>

  1. Substitute in variables:                                                                                 \displaystyle 8 = \frac{5(1)^4}{4} + \frac{9}{1} + C
  2. Evaluate exponents:                                                                                     \displaystyle 8 = \frac{5}{4} + \frac{9}{1} + C
  3. Divide:                                                                                                           \displaystyle 8 = \frac{5}{4} + 9 + C
  4. Add:                                                                                                               \displaystyle 8 = \frac{41}{4} + C
  5. [Subtraction Property of Equality] Isolate <em>C</em>:                                               \displaystyle \frac{-9}{4} = C
  6. Rewrite:                                                                                                          \displaystyle C = \frac{-9}{4}

Particular Solution: \displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}

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

Unit: Differentials Equations and Slope Fields

Book: College Calculus 10e

3 0
3 years ago
16,900,000 in scientific notation
stich3 [128]
16,900,000 in scientific notation is 16.9 times 10^7.

When you move the decimal point to the 1, it goes 7 spaces to the left, so the exponent will be a positive 7.
6 0
3 years ago
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