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grandymaker [24]
3 years ago
5

PLEASE HELPPP, I’ll brainliest the first person to answer!

Mathematics
1 answer:
AlexFokin [52]3 years ago
7 0

Answer: 3

Step-by-step explanation:

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Provide a counterexample for the proposition: If n > 3 then n> 5.
MariettaO [177]

Answer:

Your answer would be C: 4

Step-by-step explanation:

Well, I mean, 4 is greater than 3, but is less than 5. Which proves the n > 5 part wrong.

3 0
3 years ago
Can someone help me with these two questions?
fenix001 [56]

Answer:

17) A(-1,1) B(-1,4) C(5,1)

18) (From origin moving rightwards)

     A(0,0) B(a, 0) C(a,a) D(0,a)

Step-by-step explanation:

17) You just count the steps, how many steps left, right, up, or down

18) The first point is at the origin, so 0,0

    The second point that lies on the x axis is (a,0) as it is a distance from the origin, 0 because it's on the x-axis.

     The third point is (a,a) because it is a distance to the right of the origin AND a distance upwards from the x axis.

      The fourth point (0,a) because it is on the vertical line (0) and is a distance above the origin.

5 0
3 years ago
Transformations reflection practice
SOVA2 [1]

Answer:

The Reflection:

<em>'T</em>(-2,-2), <em>'C</em>(-2,-5), <em>'Z</em>(-5,-4), <em>'F</em>(-5,0)

5 0
3 years ago
13. Simplify –3√ 2 + 3√ 8 A. 0 B. 3√ 2 C. 9√ 2 D. –3√ 2
valina [46]

Answer:

B. 3√2.

Step-by-step explanation:

  1. -3√2 + 3√4x2
  2. -3√2 + 6√2
  3. 3√2

These are all of the steps to completely, find the correct answer to your question.

Hope this helps!!!

Kyle.

7 0
3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
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