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KIM [24]
2 years ago
5

Hi. I need help wih this question (see image). Please show workings.​

Mathematics
1 answer:
Kipish [7]2 years ago
4 0

Answer:

\frac{dy}{dx} = 12x³ - 6x² - 7

Step-by-step explanation:

Differentiate each term using the power rule

\frac{d}{dx} (ax^{n} ) = nax^{n-1}

Given

y = 3x^{4} - 2x³ - 7x + 5, then

\frac{dy}{dx} = 4.3x^{4-1} - 3.2x^{3-1} - 7x^{1-1} + 0

   = 12x³ - 6x² - 7

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Find the limit as x approaches 0 for cos(2x) / x
zhenek [66]

We are asked to determine the limits of the function cos(2x) / x as x approaches to zero. In this case, we first substitute zero to x resulting to 1/0.  A number, any number divided by zero is always equal to infinity, Hence there are no limits to this function.
5 0
3 years ago
Which is an asymptofe of the graph the function y=cot(x-2π/3)
Shalnov [3]

Hello.


C) x=4π/3


The variable x in the cotangent argument has a unit coefficient, so the period is π, just as it is in the parent function cot(x).


Can you graph y = cot(x)? By subtracting the constant π/6 from the argument, that graph is translated to the right by π/6. Just as with cot(x), it is decreasing everywhere.


Have a nice day



5 0
3 years ago
Consider an experiment with sample space S 5 50, 1, 2, 3, 4, 5, 6, 7, 8, 96 and the events A 5 {0, 2, 4, 6, 8} B 5 {1, 3, 5, 7,
Brut [27]

Answer with Step-by-step explanation:

S={0,1,2,3,4,5,6,7,8,9}

A={0,2,4,6,8}

B={1,3,5,7,9}

C={0,1,2,3,4}

D={5,6,7,8,9}

a.A'=S-A

A'={0,1,2,3,4,5,6,7,8,9}-{0,2,4,6,8}

A'={1,3,5,7,9}

b.C'=S-C

C'={0,1,2,3,4,5,6,7,8,9}-{0,1,2,3,4}

C'={5,6,7,8,9}

c.D'=S-D

D'={0,1,2,3,4,5,6,7,8,9}-{5,6,7,8,9}

D'={0,1,2,3,4}

d.A\cup B={0,2,4,6,8}\cup{1,3,5,7,9}

A\cup B={0,1,2,3,4,5,6,7,8,9}=S

e.A\cup C={0,2,4,6,8}\cup{0,1,2,3,4}

A\cup C={0,1,2,3,4,6,8}

f.A\cup D={0,2,4,6,8}\cup{5,6,7,8,9}

A\cup D={0,2,4,5,6,7,8,9}

8 0
3 years ago
Please help, i have no idea how to do this
notsponge [240]

Answer:

  8 square units

Step-by-step explanation:

The figure is a trapezoid. The area of it is given by the formula ...

  A = (1/2)(b1 +b2)h

where b1 and b2 are the lengths of the parallel bases and h is the distance between them.

Your figure shows the base lengths to be 5 and 3, and their separation to be 2. Filling the numbers in the formula, we have ...

  A = (1/2)(5 +3)(2) = (1/2)(8)(2) = 4·2 = 8

The area of the figure is 8 square units.

_____

The right-pointing arrows on the horizontal lines identify those lines as being parallel. The right-angle indicator and the 2 next to the dotted line indicate the perpendicular distance between the parallel lines is 2 units.

4 0
3 years ago
If anyone can check my work for me that would be greatly appreciated I'm having trouble.
krek1111 [17]
The answer is -13. Solution: = |-4b - 8| + |-1 - b^2| + 2b^3 = |-4(-2) - 8| + |-1 - -2^2| + 2(-2)^3 = |8-8| + |-1+4| + 2(-8) = |0| + |3| + (-16) = -13
8 0
2 years ago
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