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Rashid [163]
3 years ago
13

Match each equation with the correct property. 5 + 6 = 6 + 5 2(5 + 6) = 2(5) + 2(6) (3 x 5) x 6 = 3 x (5 x 6) 5 x 6 = 6 x 5

Mathematics
2 answers:
STALIN [3.7K]3 years ago
5 0

Answer:

It is x2=x

Step-by-step explanation:

amid [387]3 years ago
5 0
The answer is x2=x how this helped!
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I need help!! Describe how to determine the average rate of change between x = 1 and x = 3 for the function f(x) = 3x^3 + 1. Inc
Lostsunrise [7]

Answer:

39

Step-by-step explanation:

6 0
3 years ago
If the point A (6,9) is translated 4 units left, then find the new coordinates of the point A.
Alex_Xolod [135]

Answer:

(2, 9 )

Step-by-step explanation:

A translation of 4 units left is equivalent to subtracting 4 from the value of the x- coordinate, that is

(6, 9 ) → (6 - 4, 9 ) → (2, 9 )

6 0
3 years ago
2. Find the midpoint of the segment with endpoints (1,4), (3,6)
I am Lyosha [343]

Answer:

2,5

Step-by-step explanation:

(1+3)/2, (4+6)/2

5 0
3 years ago
How to differentiate ?
Bas_tet [7]

Use the power, product, and chain rules:

y = x^2 (3x-1)^3

• product rule

\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}

• power rule for the first term, and power/chain rules for the second term:

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}

• power rule

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3

Now simplify.

\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) =  \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)

Differentiate both sides and you end up with the same derivative:

\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2

7 0
2 years ago
David made
Zina [86]
Hi there

You made per hour
342÷18
=$19
You made for 8 hours
$19×8
=$152

Hope it helps
5 0
3 years ago
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