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3 years ago

Find the range of the function F(x) = (x-2)2 +2

Mathematics

1 answer:

viktelen [127]3 years ago

8 0

(x - 2)^2 will always be positive and will have a minimum value of 0

so f(x) will have minimum of 2

Range is [2,∞)

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I need help simplifying this

abruzzese [7]

Answer:

x^(1/12).

Step-by-step explanation:

By the laws of indices:

(x ^ 1/2)^1/6)

= x ^(1/2*1/6)

= x^(1/12).

5 0

3 years ago

Find the fourth roots of 16(cos 200° + i sin 200°).

NeTakaya

Answer:

See below.

Step-by-step explanation:

To find roots of an equation, we use this formula:

$z^{\frac{1}{n}}=r^{\frac{1}{n}}(cos(\frac{\theta}{n}+\frac{2k\pi}{n} )+\mathfrak{i}(sin(\frac{\theta}{n}+\frac{2k\pi}{n})),$ where k = 0, 1, 2, 3... (n = root; equal to n - 1; dependent on the amount of roots needed - 0 is included).

In this case, n = 4.

Therefore, we adjust the polar equation we are given and modify it to be solved for the roots.

Part 2: Solving for root #1

To solve for root #1, make k = 0 and substitute all values into the equation. On the second step, convert the measure in degrees to the measure in radians by multiplying the degrees measurement by $\frac{\pi}{180}$ and simplify.

$z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(0)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(0)\pi}{4}))$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))$

$z^{\frac{1}{4}} = 2(sin(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))$

Root #1:

$\large\boxed{z^\frac{1}{4}=2(cos(\frac{19\pi}{36}))+\mathfrack{i}(sin(\frac{19\pi}{38}))}$

Part 3: Solving for root #2

To solve for root #2, follow the same simplifying steps above but change k to k = 1.

$z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(1)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(1)\pi}{4}))$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{2\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{2\pi}{4}))\\$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{2}))\\$

Root #2:

$\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{7\pi}{9}))+\mathfrak{i}(sin(\frac{7\pi}{9}))}$

Part 4: Solving for root #3

To solve for root #3, follow the same simplifying steps above but change k to k = 2.

$z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(2)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(2)\pi}{4}))$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{4\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{4\pi}{4}))\\$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\pi))+\mathfrak{i}(sin(\frac{5\pi}{18}+\pi))\\$

Root #3:

$\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{23\pi}{18}))+\mathfrak{i}(sin(\frac{23\pi}{18}))}$

Part 4: Solving for root #4

To solve for root #4, follow the same simplifying steps above but change k to k = 3.

$z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(3)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(3)\pi}{4}))$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{6\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{6\pi}{4}))\\$

$z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{3\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{3\pi}{2}))\\$

Root #4:

$\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{16\pi}{9}))+\mathfrak{i}(sin(\frac{16\pi}{19}))}$

The fourth roots of 16(cos 200° + i(sin 200°) are listed above.

3 0

3 years ago

Which is the independent variable in this graph?

weqwewe [10]

The population because it’s what’s changing

4 0

2 years ago

Read 2 more answers

Explain how you might go from gallons to cups using more than one conversion factor.

damaskus [11]

Answer:

because it’s different system

Step-by-step explanation: When going from the metric system to another, You do conversion factor.

8 0

2 years ago

Read 2 more answers

Can someone show me how to work out this example? (5+44+182−20−10)÷(2+5)

lora16 [44]

Answer:

approximately 35.9

Step-by-step explanation:

(5+44+182-20-10)÷(2+5)

(99+182-20-10)÷(7)

(281-20-10)÷(7)

(261-10)÷÷(7)

(251)÷(7)

approximately 35.9

6 0

3 years ago