factor the polynomial. 17xy^4 51x^2y^3 a) 7xy (10y^4 44xy^2) b) 17xy^3 (y 3x) c) 17xy (1 1/3xy^2) d) xy (17 51xy^2)

nirvana33 [79]

Hey there! :D

m-1/6-4m+5/6

m-4m=-3m

-1/6+5/6= 4/6

-3m+4/6 <== simplified expression

I hope this helps!
~kaikers

7 0

3 years ago

If a problem is 8(2x+4) = 2(1x+3) is the first step done of PEMDAS for THIS SPECIFIC PROBLEM parentheses or multiplication?

garri49 [273]

The first step is the parenthesis

<h3>How to determine the first step?</h3>

The equation is given as:

8(2x+4) = 2(1x+3)

Using PEMDAS, the first step is to open the brackets by multiplication.

So, we have:

16x + 32 = 2x + 6

Hence, the first step is the parenthesis

Read more about expressions at:

brainly.com/question/723406

#SPJ1

3 0

2 years ago

Which numbers belong to the solution set of the inequality?

grandymaker [24]

Answer: B

Step-by-step explanation:

x+9<78 => x< 78-9=> x< 69 => b right

6 0

3 years ago

Read 2 more answers

Help me please!!! Jdndkdjdnendnd

BartSMP [9]

Answer:

To graph a parabola, use the coefficient a and coefficient b values from your parabolic equation in the formula x = -b ÷ 2a to solve for x, which is the first coordinate of the vertex. Next, plug x back into your equation to solve for y, which is the second coordinate of the vertex.

i dont know if it is correct though.

3 0

3 years ago

Y=sqrt(x)(8x-5) find the derivative

garri49 [273]

Answer:

$\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

General Formulas and Concepts:

Algebra I

Exponentials [Fractions] - Are radicals
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \sqrt{x}(8x - 5)$

Step 2: Differentiate

$\displaystyle f(x) = \sqrt{x}, \ g(x) = (8x - 5)$

Product Rule: $\displaystyle y' = \frac{d}{dx}[\sqrt{x}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Rewrite: $\displaystyle y' = \frac{d}{dx}[x^{\frac{1}{2}}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{1 - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{0}$
Rewrite: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8$
Multiply: $\displaystyle y' = \frac{8x + 5}{2x^{\frac{1}{2}}} + 8\sqrt{x}$
Rewrite: $\displaystyle y' = \frac{8x + 5}{2\sqrt{x}} + 8\sqrt{x}$
Add/Rewrite: $\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

3 0

3 years ago