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3 years ago

If an average grade of an English class rise from 70 to 85 what is the approximate percent increase

Mathematics

1 answer:

Bogdan [553]3 years ago

4 0

Answer:

15% IS THE ANSWER

Step-by-step explanation:

You might be interested in

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

Paraplatin 360mg/m2 in 150ml normal saline via ice pump. Drug run over 30 minutes. At what rate should the nurse at IV pump

olasank [31]

Drag concentration( 360mg) per body surface (m2) is irrelevant to the infusion rate, which is volume divided by time, i.e. 150/(30/60) = 300ml/h

7 0

3 years ago

Scott started his banking account with $150 and is spending $7 per day on lunch what would the x axis be labeled

Mariulka [41]

X: number of days
y: money available

8 0

3 years ago

Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

statuscvo [17]

9514 1404 393

Answer:

(a) x = (3 -ln(3))/5 ≈ 0.819722457734

(b) y = 10

Step-by-step explanation:

(a) Taking the natural log of both sides, we have ...

2x +1 = ln(3) +4 -3x

5x = ln(3) +3 . . . . . . . . add 3x-1

x = (ln(3) +3)/5 ≈ 0.820

(b) Assuming "lg" means "log", the logarithm to base 10, we have ...

log(y -6) +log(y +15) = 2

(y -6)(y +15) = 100 . . . . . . . taking antilogs

y^2 -9x -190 = 0 . . . . . . . . eliminate parentheses, subtract 100

(y -19)(y +10) = 0 . . . . . . . . factor

The values of y that make these factors zero are -19 and 10. We know from the first term that (y-6) > 0, so y > 6. That means y = -19 is an extraneous solution.

The solution is ...

y = 10

When solving equations using a graphing calculator, it often works well to define a function f(x) such that the solution is f(x) = 0, the x-intercept(s). That form is easily obtained by subtracting the right side of the equation from both sides of the equation. In part (a) here, that is ...

f(x) = e^(2x+1) -3e^(4-3x)