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3 years ago

State the property of addition that justifies each numbered step in the following simplification.

Mathematics

1 answer:

Ierofanga [76]3 years ago

4 0

Well you would use PEMDAS solve what's in the parenthesis first and the add that to 56 and you should get your answer.

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Parabola,Can someone please help me please ?

ivanzaharov [21]

Answer:

a. opens: downward

b. vertex: (4, 2)

c. x-intercepts: (2, 0), (6, 0); y-intercept: (0, -6)

d. a.o.s.: x = 4

Step-by-step explanation:

Based on the picture the parabola opens downwards. Right away you know that the "a" value in the parabola's equation will be negative.

To find the coordinates of the vertex, you'll need to look at the picture. Remember: look on the x-axis first and then the y-axis next.

The very top of the parabola (maximum point = vertex) is at the x-value of 4 and the y-value of 2. This means the coordinates of the vertex of this parabola is (4, 2).

To find the x-intercepts, look at the x-axis and see where the parabola crosses it. The parabola goes through the x-axis at the x-values of 2 and 6, so the x-intercepts of this parabola are (2, 0) and (6, 0).

To find the y-intercept, look at the y-axis and see where the parabola crosses it. The parabola intersects the y-axis at the y-value of -6, so (0, -6) would be the y-intercept.

The equation for the axis of symmetry of a graph is always x = h, where h is the x-value of the coordinate of the vertex.

Since the vertex is (4, 2), where 4 = h, the equation of the axis of symmetry would be x = 4.

4 0

4 years ago

F(x) = Square root of x plus eight.; g(x) = 8x - 12 Find f(g(x))

Liono4ka [1.6K]

$f(x)= \sqrt{x+8} \Rightarrow f(g(x))= \sqrt{(8x-12)+8} = \sqrt{8x -4}$

7 0

3 years ago

Read 2 more answers

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

garri49 [273]

Answer:

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

General Formulas and Concepts: <u> </u>

<u>Algebra I</u>

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

<u>Calculus</u>

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:

<u>Step 1: Define</u>

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

<u>Step 2: Differentiate</u>

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

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