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2 years ago

Last two!!(ty for all the help) :

2 answers:

7 0

With the b2+b- 12 your gonna want to do this
b2+b-12
b2+4b-3b-12
which is the sum product
after doing that you wanna common the factors from the two pairs. Which is b2+4b-3b-12 then you do you parentheses around b(b+4)-3(b+4) just like that after you do that then you rewrite it in factored form b(b+4)-3(b+4) you then want to rewrite it to this (b-3)(b+4) after that your solution will be (b-3)(b+4) for the first one.

scoray [572]2 years ago

6 0

(B-3)(b+4) is the answer to the solution

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A total of 50 people play darts each players given three darts to try to hit the bull's-eye 21 at the darts hit the bull's-eye t

Yuliya22 [10]

Answer:

42%

Step-by-step explanation:

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3 years ago

A music store owner noted that CD sales had dropped 33% from one quarter to the next. If the owner sold 579 units in the first q

Paul [167]

Answer: 387.93 units

Step-by-step explanation:

You have the following information given in the problem above:

- The sales had dropped 33% from one quarter to the next.

- The owner sold 579 units in the first quarter.

Thefore, to calculate the number of units she sold in the next quarter (which you can call x), you must can apply the following proccedure:

$x=579units-(0.33)(579units)$

Then the answer is the following:

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4 0

3 years ago

Please help it and show all the work please help

kompoz [17]

Answer:

$\sqrt[n]{x} => n \ is \ the \ index \ and \ x \ is \ the \ radicand\\\\1. b) -23^7\\\\2. a) 2\\\\3. b) \sqrt{3} \cdot\sqrt{2} =\sqrt{6}$

For q1 and q2 I have given explanation already.

q3 to multiply radicals the index must be same.

option a . different index. ∛5 √2

option b . same index so √3 √2 = √6

option c. again different index.

7 0

3 years ago

Hi, how do we do this question?

Nutka1998 [239]

Answer:

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$

General Formulas and Concepts:

Algebra I

Terms/Coefficients
Factoring

Algebra II

Polynomial Long Division

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals
Integration Constant C
Indefinite Integrals

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

Logarithmic Integration

U-Substitution

Step-by-step explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

Step 1: Define

Identify

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx$

Step 2: Integrate Pt. 1

[Integrand] Rewrite [Polynomial Long Division (See Attachment)]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$
[1st Integral] Reverse Power Rule: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 3x + 1$
[u] Differentiate [Basic Power Rule]: $\displaystyle du = 3 \ dx$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx$
[Integral] U-Substitution: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du$
[Integral] Logarithmic Integration: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C$
Back-Substitute: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C$
Factor: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3} \bigg) + C$
Rewrite: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$