All categories

3 years ago

Which of these graphs represents a function?

Mathematics

1 answer:

Naddik [55]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

Write an expression equivalent to `b+b+b+b+b` that is a product of a coefficient and a variable.

mestny [16]

Answer:

b^5

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

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

a family has five children. the probability of having a girl is 1/2. whats probability of having at leasr 4 girls g'

Dima020 [189]

Answer:

Probability of having at least 4 Girls

= 0.6875

Step-by-step explanation:

Probability of having at least 4 Girls is 1-probability of having exactly 3 girls

Total number of children= 5 = N

Probability of having a girl p = 0.5

Probability of not having a girl q= 0.5

X= 3

Probability of at least 4 girls is given by

Probability= NCX(p)^x(q)^(N-x)

Probability = 5C3(0.5)^3(0.5)^(5-3)

Probability = 5C3(0.5)^3(0.5)^2

Probability= 5!/3!2!(0.5)^3(0.5)^2

Probability= 10(0.125)(0.25)

Probability= 0.3125

Probability of having at least 4 Girls

= 1- 0.3125

= 0.6875

7 0

3 years ago

The table below shows the cube roots of different numbers: Number (x) 27 −27 125 −125 Cube root of the number (y) 3 −3 5 −5

kolbaska11 [484]

Part A

The table does represent y as a function of x because each input (values of x) related to exactly one output. Refer to the diagram below.

Part B

We have
f(x) = 925x + 1000
f(24) = 925 (24) + 1000
f(24) = 23200

f(24) represent the value of renting for 24 months

5 0

3 years ago

Arthur earned $136 in three weeks . He goes back to school in one more week . He needs at least $189 to buy the new coat that he

nadya68 [22]

He must earn 53 dollars

7 0

3 years ago

Read 2 more answers