Summary & pratice sheets.

How much further up the wall does the ladder in figure 2 reach than

ASHA 777 [7]

The ladder in figure 2 reaches 3.2 feet further up than the ladder in figure 1.

Step-by-step explanation:

Step 1:

In both the given figures, the ladder, the wall and the floor form a right-angled triangle. The floor is the adjacent side, the wall is the opposite side and the ladder is the hypotenuse.

According to the Pythagorean theorem,

$a^{2}+b^{2}=c^{2}$ , where c is the length of the hypotenuse while a and b are the lengths of the other two sides.

Step 2:

For the ladder in figure 1, assume the distance from the floor to the ladder's top is x feet. So a = x, b = 8 and c = 10 (hypotenuse).

$a^{2}+b^{2}=c^{2}$ , $x^{2}+8^{2}=10^{2}$ , $x^{2}=100-64=36, x=\sqrt{36}=6 \text { feet }$ .

So the distance between the floor and the ladder's top is 6 feet.

Step 3:

For the ladder in figure 2, assume the distance between the floor and the ladder's top is y feet. So a = y, b = 4 and c = 10 (hypotenuse).

$a^{2}+b^{2}=c^{2}$ , $y^{2}+4^{2}=10^{2}$ , $y^{2}=100-16=48, x=\sqrt{84}=9.1651 \text { feet }$ .

So the distance between the floor and the ladder's top is 9.1651 feet.

Step 4:

The difference in heights = The wall height in figure 2 - the wall height in figure 1.

The difference in heights = 9.1651 feet - 6 feet = 3.1651 feet.

Rounding this off to the nearest tenth of a foot, we get 3.2 feet.

4 0

3 years ago

The cheerleader section is on the basketball court. Each row requires a section of floor that is 1 ¾ yards by 3 ⅓ yards. How man

Lady_Fox [76]

Answer:

The area of the floor space is equal to 5.83 sq yards.

Step-by-step explanation:

Given that,

Each row requires a section of floor that is 1 ¾ yards by 3 ⅓ yards.

We need to find how many square yards of floor space are taken up by one row of cheerleaders.

We know that, the area of a rectangle is given by :

A = lb

So,

$A=\dfrac{7}{4}\times \dfrac{10}{3}\\\\A=5.83\ yards^2$

So, the area of the floor space is equal to 5.83 sq yards.

5 0

3 years ago

Margaret is playing a game with a standard deck of 52 cards which contains 4 jacks. The first player to select a jack gets to st

Lelechka [254]

Answer:

Step-by-step explanation: The theoretical probability that Margaret, when choosing first, will choose a jack, is 1/13.

welcomeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

5 0

3 years ago

Read 2 more answers

You score 5 goals. Your friend scores

EastWind [94]

Answer:

11 goals

Step-by-step explanation:

Me: 5 goals

Friends: 5+1 = 6 goals

Me + friends = 5 goals + 6 goals = 11 goals

7 0

1 year ago

Read 2 more answers

Need help please its Calculus. Ill give the 5 stars as well.

algol13

Answer:

$\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$

General Formulas and Concepts:

Pre-Algebra

Order of Operations
Equality Properties

Algebra I

Functions
Function Notation
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Algebra II

Natural logarithms ln and Euler's number e

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Slope Fields

Separation of Variables

Solving Differentials

Integrals

Antiderivatives

Integration Constant C

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

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

U-Substitution

Logarithmic Integration: $\displaystyle \int {\frac{1}{u}} \, dx = ln|u| + C$

Step-by-step explanation:

*Note:

When solving differential equations in slope fields, disregard the integration constant C for variable y.

Step 1: Define

$\displaystyle \frac{dy}{dx} = x^2(y - 1)$

$\displaystyle f(0) = 3$

Step 2: Rewrite

Separation of Variables. Get differential equation to a form where we can integrate both sides and rewrite Leibniz Notation.

[Separation of Variables] Rewrite Leibniz Notation: $\displaystyle dy = x^2(y - 1) \ dx$
[Separation of Variables] Isolate y's together: $\displaystyle \frac{1}{y - 1} \ dy = x^2 \ dx$

Step 3: Find General Solution Pt. 1

[Differential] Integrate both sides: $\displaystyle \int {\frac{1}{y - 1}} \, dy = \int {x^2} \, dx$
[dx Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle \int {\frac{1}{y - 1}} \, dy = \frac{x^3}{3} + C$

Step 4: Find General Solution Pt. 2

Identify variables for u-substitution for dy.

Set: $\displaystyle u = y - 1$
Differentiate [Basic Power Rule]: $\displaystyle du = dy$

Step 5: Find General Solution Pt. 3

[dy Integral] U-Substitution: $\displaystyle \int {\frac{1}{u}} \, du = \frac{x^3}{3} + C$
[dy Integral] Integrate [Logarithmic Integration]: $\displaystyle ln|u| = \frac{x^3}{3} + C$
[Equality Property] e both sides: $\displaystyle e^\bigg{ln|u|} = e^\bigg{\frac{x^3}{3} + C}$
Simplify: $\displaystyle |u| = Ce^\bigg{\frac{x^3}{3}}$
Rewrite: $\displaystyle u = \pm Ce^\bigg{\frac{x^3}{3}}$
Back-Substitute: $\displaystyle y - 1 = \pm Ce^\bigg{\frac{x^3}{3}}$
[Equality Property] Isolate y: $\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1$

General Form: $\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1$

Step 6: Find Particular Solution

Substitute in function values [General Form]: $\displaystyle 3 = \pm Ce^\bigg{\frac{0^3}{3}} + 1$
Simplify: $\displaystyle 3 = \pm C + 1$
[Equality Property] Isolate C: $\displaystyle 2 = \pm C$
Rewrite: $\displaystyle C = 2$
Substitute in C [General Form]: $\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$

∴ our particular solution is $\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$ .

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

Unit: Differentials and Slope Fields

Book: College Calculus 10e

6 0

3 years ago