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3 years ago

A middle school needs buses to

Mathematics

2 answers:

Tresset [83]3 years ago

7 0

Answer:

13 buses ml hope that helps

cricket20 [7]3 years ago

4 0

Answer:

13 buses

Step-by-step explanation:

There are 579 students and each bus carries 48 students so:

579 ÷ 48 = 12.1

Now you could say that only 12 buses are needed but there would be an extra student remaining so you need 13 buses.

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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:

Pre-Algebra

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

Algebra I

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

Calculus

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:

Step 1: Define

F: y = √(15 - x)

G: y = √(15 - 3x)

H: y = 0

Step 2: Find Bounds of Integration

Solve each equation for the x-value for our bounds of integration.

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

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

Step 3: Find Area of Region

Integration Part 1

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]: $\displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx$

Step 4: Identify Variables

Set variables for u-substitution for both integrals.

Integral 1:

u = 15 - x

du = -dx

Integral 2:

z = 15 - 3x

dz = -3dx

Step 5: Find Area of Region

Integration Part 2

[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

Please helpp!! I need help placing and label the following numbers on the number line

Alchen [17]

-1 goes on the number that is -1.
1.75 goes on the 3rd line in between 1 and 2.
-2 goes on the number -2.
-2 1/2 on the 2nd line after 2.
-5/2 or -2 1/2 goes on the 2 line before -2.
And 9/4 or 2 1/4 goes on goes on the 1st line after 2.

Hope this helps!! :)

3 0

3 years ago

Solve compound inequality x ≥ -3 or x < 3

Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

Compound Inequalities is a method to find the variable value for the given compound inequalities. The two inequalities would be joined by the word 'and' or 'or'. It is presented in a number line. When the arrow points away from each other in the number line, then the inequalities are joined by 'and', if arrow points on same direction ...

5 0

3 years ago

The perimeter of a particular square and the circumference of a particular circle are equal. What is the ratio of the area of th

Tasya [4]

Ratio of the area of the square to the area of the circle is $\frac{\pi}{4} : 1$

In terms of fraction $\frac{\pi}{4}$

Solution:

Given that perimeter of a particular square and the circumference of a particular circle are equal

To find: ratio of the area of the square to the area of the circle

Let the side of square be "a" and radius of circle be "r"

The perimeter of square is given as:

Perimeter of square = 4a

The circumference of circle is given as:

Circumference of circle = 2πr

From given,

perimeter of a particular square and the circumference of a particular circle are equal

perimeter of a particular square = circumference of a particular circle

$4a = 2 \pi r\\\\a = \frac{\pi r}{2}$

We need to find the ratio of the area of the square to the area of the circle.

We know that,

$\text{ Area of the square }= a^2\\\\\text{ Area of the circle }= \pi r^2$

Ratio of area of square to area of circle = $a^2 : \pi r^2$

Substituting the value of $a = \frac{\pi r}{2}$ in above ratio,

Ratio of area of square to area of circle = $(\frac{\pi r}{2})^2 : \pi r^2$

On reducing to lowest terms we get,

⇒ $\frac{\pi^2 r^2}{4} : \pi r^2 = \frac{\pi}{4} : 1$

In terms of fraction we get,

$\frac{\pi}{4} : 1 = \frac{\frac{\pi}{4}}{1} = \frac{\pi}{4}$

Thus ratio of the area of the square to the area of the circle is $\frac{\pi}{4} : 1$

3 0

4 years ago

Please simplify will give brainliyest

dimulka [17.4K]

$\displaystyle\bf\\-\sqrt{m^4n^7}=\\\\=-\sqrt{m^{2\times2}\times n^{3\times2+1}}=\\\\=-\sqrt{m^{2\times2}\times n^{3\times2}\times n}}=\\\\=-\sqrt{\Big(m^2\Big)^2\times \Big(n^3\Big)^2\times n}}=\\\\=-\sqrt{\Big(m^2\times n^3\Big)^2\times n}}=\\\\=\boxed{\bf-m^2n^3\sqrt{n}}$

4 0

4 years ago