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3 years ago

11

help! I need an answer to this question, I also need to know HOW to solve it, so that I'm able to do the other questions on my o

wn. brainliest if you get it right

2 answers:

soldier1979 [14.2K]3 years ago

6 0

Answer:

What grade?

Step-by-step explanation:

kondor19780726 [428]3 years ago

5 0

Answer: pie 3/2

Step-by-step explanation:

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Three more than the product of a number and 2 equals 7.<br> for the unknown number

Alchen [17]

3 + 2n = 7

Hope this helps and have a nice day/night!!!

7 0

3 years ago

Read 2 more answers

Sean asked 10 of his classmates the amount of their weekly allowance. He recorded the information in this table. $15 $15 $22 $14

Anarel [89]

The interquartile range = Third quartile - First quartile = 22 - 15 = 7

The range = 25 - 12 = 13

<h3>What is the Interquartile Range (IQR) and Range ofa Data?</h3>

Range = largest data value - the least data value

Interquartile range = Q3 - Q1

Given the following weekly allowance of Sean's classmates as, 15, 15, 22, 14, 12, 24, 16, 25, 15, 15:

Third quartile (Q3) = 22

First quartile (Q1) = 15

Interquartile range = 22 - 15 = 7

Range = 25 - 12 = 13

Learn more about trhe interquartile range on:

brainly.com/question/4102829

#SPJ1

8 0

2 years ago

Find the area of the region enclosed by the graphs of these equations. (CALCULUS HELP)

sergiy2304 [10]

Answer:

$\displaystyle A = \frac{20\sqrt{15}}{3}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
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>

Terms/Coefficients
Graphing
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Area - Integrals

U-Substitution

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]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

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

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

<u>Step 1: Define</u>

F: y = √(15 - x)

G: y = √(15 - 3x)

H: y = 0

<u>Step 2: Find Bounds of Integration</u>

<em>Solve each equation for the x-value for our bounds of integration.</em>

F

Set <em>y</em> = 0: 0 = √(15 - x)
[Equality Property] Square both sides: 0 = 15 - x
[Subtraction Property of Equality] Isolate <em>x</em> term: -x = -15
[Division Property of Equality] Isolate <em>x</em>: x = 15

G

Set y = 0: 0 = √(15 - 3x)
[Equality Property] Square both sides: 0 = 15 - 3x
[Subtraction Property of Equality] Isolate <em>x</em> term: -3x = -15
[Division Property of Equality] Isolate <em>x</em>: x = 5

This tells us that our bounds of integration for function F is from 0 to 15 and our bounds of integration for function G is 0 to 5.

We see that we need to subtract function G from function F to get our area of the region (See attachment graph for visual).

<u>Step 3: Find Area of Region</u>

<em>Integration Part 1</em>

Rewrite Area of Region Formula [Integration Property - Subtraction]: $\displaystyle A = \int\limits^b_a {f(x)} \, dx - \int\limits^d_c {g(x)} \, dx$
[Integral] Substitute in variables and limits [Area of Region Formula]: $\displaystyle A = \int\limits^{15}_0 {\sqrt{15 - x}} \, dx - \int\limits^5_0 {\sqrt{15 - 3x}} \, dx$
[Area] [Integral] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx$

<u>Step 4: Identify Variables</u>

<em>Set variables for u-substitution for both integrals.</em>

Integral 1:

u = 15 - x

du = -dx

Integral 2:

z = 15 - 3x

dz = -3dx

<u>Step 5: Find Area of Region</u>

<em>Integration Part 2</em>

[Area] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = -\int\limits^{15}_0 {-(15 - x)^{\frac{1}{2}}} \, dx + \frac{1}{3}\int\limits^5_0 {-3(15 - 3x)^{\frac{1}{2}}} \, dx$
[Area] U-Substitution: $\displaystyle A = -\int\limits^0_{15} {u^{\frac{1}{2}}} \, du + \frac{1}{3}\int\limits^0_{15} {z^{\frac{1}{2}}} \, dz$
[Area] Reverse Power Rule: $\displaystyle A = -(\frac{2u^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15} + \frac{1}{3}(\frac{2z^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15}$
[Area] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = -(-10\sqrt{15}) + \frac{1}{3}(-10\sqrt{15})$
[Area] Multiply: $\displaystyle A = 10\sqrt{15} + \frac{-10\sqrt{15}}{3}$
[Area] Add: $\displaystyle A = \frac{20\sqrt{15}}{3}$

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

Unit: Area Under the Curve - Area of a Region (Integration)

Book: College Calculus 10e

3 0

3 years ago

5÷x-4 - 2÷x+2 = 4÷x-4

Alborosie

I'd suggest you begin by subtracting 4 / (x-4) from both sides. Doing that would leave you with 1 / (x-4) - 2 / (x+2) = 0.

LCD is (x-4)(x+2). Mult. all three terms by (x-4)(x+2). The resulting equation is

x+2 - 2(x-4) = 0. Then x+2 = 2x - 8 => x = 10

Subst. 10 for x in the original equation to verify that 10 is indeed a solution.

8 0

3 years ago

Please help me with number 19

eimsori [14]

The new equation will be (x-1)² + (y-18)²=36

2) x²+y²+6x+12y=4
(x²+6x+?)-(?) + (y²+12y +??) -(??)=4

(x²+6x+9)-(9) + (y²+12y +36) -(36)=4
(x+3)² - 9 + (y+6)² - 36 = 4

(x+3)² +(y+6)² = 49

6 0

3 years ago