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3 years ago

What is the distance between the points (2, 1) and (14, 6) on a coordinate

Mathematics

2 answers:

Naddika [18.5K]3 years ago

8 0

The distance between the points (2,1) and (14,6) on a coordinate is C. 13 units

DochEvi [55]3 years ago

6 0

Answer:12/5

Step-by-step explanation:

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Which equation below represents a line with a slope of - 2/3 that passes through the point (-4.5 , 10)?

Lostsunrise [7]

Answer:

Option C

Step-by-step explanation:

If a line passes through a point, then that point is called a solution to the line's equation. Substituting the x and y values of that solution into the equation will give a true statement. So, to find out which option is correct, we can substitute the x and y values of (-4.5, 10) into each equation and see if the result is a true statement.

Let's try this with option C. To make things easier, convert -4.5 into decimal form: $-4\frac{1}{2}$ . Substitute $-4\frac{1}{2}$ for x and 10 for y in the equation, then solve:

$y = -\frac{2}{3} x+7\\\\10 = -\frac{2}{3} (-4\frac{1}{2} )+7\\10 = -\frac{2}{3} (-\frac{9}{2})+7\\10 = 3+7\\10 = 10$

10 does equal 10, so this is a true statement. Option C is the right answer.

8 0

3 years ago

Given f(x) = 4(x - 8), determine the value of f(10).

Ilya [14]

Answer:

f(10) = 8

Step-by-step explanation:

f(x) = 4(x - 8)

Let x = 10

f(10) = 4 ( 10-8)

= 4 ( 2)

= 8

8 0

3 years ago

Read 2 more answers

Please help and explain how you solved it. Thanks.

galina1969 [7]

Answer:

This is 0.14 to the nearest hundredth

Step-by-step explanation:

Firstly we list the parameters;

Drive to school = 40

Take the bus = 50

Walk = 10

Sophomore = 30

Junior = 35

Senior = 35

Total number of students in sample is 100

Let W be the event that a student walked to school

So P(w) = 10/100 = 0.1

Let S be the event that a student is a senior

P(S) = 35/100 = 0.35

The probability we want to calculate can be said to be;

Probability that a student walked to school given that he is a senior

This can be represented and calculated as follows;

P( w| s) = P( w n s) / P(s)

w n s is the probability that a student walked to school and he is a senior

We need to know the number of seniors who walked to school

From the table, this is 5/100 = 0.05

So the Conditional probability is as follows;

P(W | S ) = 0.05/0.35 = 0.1429

To the nearest hundredth, that is 0.14

3 0

3 years ago

What is the volume of the right rectangular prism below?

Vesnalui [34]

Answer:

Volume of right rectangular prism = 3.75 in³

Step-by-step explanation:

Given:

First side of rectangular prism = 2 inches

Second side of rectangular prism = 1 $\frac{1}{2}$ = 1.5 inches

Third side of rectangular prism = 1 $\frac{1}{4}$ = 1.25 inches

Find:

Volume of right rectangular prism:

Computation:

$Volume\ of\ right\ rectangular\ prism=lbh\\\\Volume\ of\ right\ rectangular\ prism=(2)(1.5)(1.25)\\\\ Volume\ of\ right\ rectangular\ prism= 3.75 in^3$

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%281-x%5E%7B2%7D%20%29%5E%7B3%2F2%7D%20%7D%20%5C%2C%20dx" id="TexFo

Ludmilka [50]

Answer: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}$ General Formulas and Concepts:

Pre-Calculus

Trigonometric Identities

Calculus

Differentiation

Derivatives
Derivative Notation

Integration

Integrals
Definite/Indefinite Integrals
Integration Constant C

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

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

U-Substitution

Trigonometric Substitution

Reduction Formula: $\displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution (trigonometric substitution).

Set u: $\displaystyle x = sin(u)$
[u] Differentiate [Trigonometric Differentiation]: $\displaystyle dx = cos(u) \ du$
Rewrite u: $\displaystyle u = arcsin(x)$

Step 3: Integrate Pt. 2

[Integral] Trigonometric Substitution: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{\frac{3}{2}} \, du$
[Integrand] Rewrite: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{\frac{3}{2}} \, du$
[Integrand] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos^4(u)} \, du$
[Integral] Reduction Formula: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{4 - 1}{4}\int \limits^a_b {cos^{4 - 2}(x)} \, dx + \frac{cos^{4 - 1}(u)sin(u)}{4} \bigg| \limits^a_b$
[Integral] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4}\int\limits^a_b {cos^2(u)} \, du$
[Integral] Reduction Formula: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg|\limits^a_b + \frac{3}{4} \bigg[ \frac{2 - 1}{2}\int\limits^a_b {cos^{2 - 2}(u)} \, du + \frac{cos^{2 - 1}(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
[Integral] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}\int\limits^a_b {} \, du + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
[Integral] Reverse Power Rule: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}(u) \bigg| \limits^a_b + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3cos(u)sin(u)}{8} \bigg| \limits^a_b + \frac{3}{8}(u) \bigg| \limits^a_b$
Back-Substitute: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(arcsin(x))sin(arcsin(x))}{4} \bigg| \limits^a_b + \frac{3cos(arcsin(x))sin(arcsin(x))}{8} \bigg| \limits^a_b + \frac{3}{8}(arcsin(x)) \bigg| \limits^a_b$
Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x)}{8} \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{\frac{3}{2}}}{4} \bigg| \limits^a_b + \frac{3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b$
Rewrite: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{\frac{3}{2}} + 3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}$

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

Unit: Integration

Book: College Calculus 10e

8 0

3 years ago

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