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3 years ago

Can u help me plz this is very hard to and idk why

Mathematics

1 answer:

-BARSIC- [3]3 years ago

7 0

The answer you are looking for is option C, ten feet!
1 yard = 3 feet, so, when we have 3 yards, it is 9 feet! Then, we add a foot making the answer 10! Hope this helps! :)

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Help! help! please help

iren2701 [21]

Answer:

to answer this i am gonna need the rise and run or the y2 -x2/ y1-x1 and then i well come back and edit this to the right answer

Step-by-step explanation:

5 0

3 years ago

How can i solve this?

kenny6666 [7]

Am sorry I try but I cant

6 0

4 years ago

103^2-102^2=_____×(103-102)=_____ pls help me...this is in factorization

Ray Of Light [21]

103^2-102^2

The property that applies here is a^2-b^2 = (a+b)(a-b)

so the answer is 103^2-102^2=(102+103)*(103-102)= numerically it is 205( as 103+102 = 205 and 103-102 = 1.....so 205*1 = 205)

The thing is that usually the factorization ends at (102+103)(102-103) ......

8 0

4 years ago

Read 2 more answers

How would you multiply (3x-8)(2x^2+4x-9)

Naily [24]

(3x-8)(2x^2+4x-9)= 6x^3−4x^2−59x+72
Hope I helped! ( Smiles )

4 0

3 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:

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