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3 years ago

Can someone help me with this question please?

Mathematics

1 answer:

Naya [18.7K]3 years ago

8 0

Bro that stuff is really hard is this rocket science

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Erik buys 2.5 pounds of cashews. If each pound of cashews costs $7.70, how much will he pay for the cashews?

Liula [17]

Answer:

$19.25

Step-by-step explanation:

2.5*7.7=19.25

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3 years ago

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Find the value of x

nataly862011 [7]

Answer:

x = 128°

Step-by-step explanation:

123 + 109 + x = 360

x = 360 - (123 + 109)

x = 360 - 232

x = 128°

7 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

Create a word problem with 3x = 42!

Delicious77 [7]

There are 42 kids in one classroom. They are all divided into 3 groups. How many kids will be in each group?

4 0

3 years ago

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-(1+7 x) – 61–7 – x)= 36

wariber [46]

Answer:

Step-by-step explanation:

-(1+7 x)=6x-61-7-x)=48x=36

8 0

3 years ago