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Andreas93 [3]
3 years ago
7

Sahir took 1 h 22 min to do his english homework 45 min to do his maths homework, and 10 min to do his urdu. How long did he tak

e to complete all his homework?
Mathematics
1 answer:
andre [41]3 years ago
6 0

Answer:

2 hr and 17 minutes

Step-by-step explanation:

add 45 to 10+= 55

55+ 1 hr 22 = 1 hour and 77 minutes which converts to:

2 hour and 17 minutes

good luck!

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Which two values of x are roots of the polynomial below? x^2-3x+5
Gemiola [76]

Given:

Polynomial x^2-3x+5

To find:

The values of x.

Solution:

x^2-3x+5=0

Quadratic equation formula:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here a=1, b=-3, c=5.

Substitute the values in the formula, we get

$x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4 \cdot 1 \cdot 5}}{2 \cdot 1}

$x=\frac{3 \pm \sqrt{9-20}}{2}

$x=\frac{3 \pm \sqrt{-11}}{2}

$x=\frac{3 - \sqrt{-11}}{2} \ \text{and} \ x=\frac{3 + \sqrt{-11}}{2}

Option E and option F are the roots of the polynomial.

The values of x are x=\frac{3 - \sqrt{-11}}{2} \ \text{and} \ x=\frac{3 + \sqrt{-11}}{2}.

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What time is 7 1/2 before 2:12 am?
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A triangle has one angle that measures 26 degrees, one angle that measures 47 degrees, and one angle that measures 107 degrees.
just olya [345]
It’s 3 and then you have a lot to go get it all
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Find all the complex roots. Write the answer in exponential form.
dezoksy [38]

We have to calculate the fourth roots of this complex number:

z=9+9\sqrt[]{3}i

We start by writing this number in exponential form:

\begin{gathered} r=\sqrt[]{9^2+(9\sqrt[]{3})^2} \\ r=\sqrt[]{81+81\cdot3} \\ r=\sqrt[]{81+243} \\ r=\sqrt[]{324} \\ r=18 \end{gathered}\theta=\arctan (\frac{9\sqrt[]{3}}{9})=\arctan (\sqrt[]{3})=\frac{\pi}{3}

Then, the exponential form is:

z=18e^{\frac{\pi}{3}i}

The formula for the roots of a complex number can be written (in polar form) as:

z^{\frac{1}{n}}=r^{\frac{1}{n}}\cdot\lbrack\cos (\frac{\theta+2\pi k}{n})+i\cdot\sin (\frac{\theta+2\pi k}{n})\rbrack\text{ for }k=0,1,\ldots,n-1

Then, for a fourth root, we will have n = 4 and k = 0, 1, 2 and 3.

To simplify the calculations, we start by calculating the fourth root of r:

r^{\frac{1}{4}}=18^{\frac{1}{4}}=\sqrt[4]{18}

<em>NOTE: It can not be simplified anymore, so we will leave it like this.</em>

Then, we calculate the arguments of the trigonometric functions:

\frac{\theta+2\pi k}{n}=\frac{\frac{\pi}{2}+2\pi k}{4}=\frac{\pi}{8}+\frac{\pi}{2}k=\pi(\frac{1}{8}+\frac{k}{2})

We can now calculate for each value of k:

\begin{gathered} k=0\colon \\ z_0=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{0}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{0}{2}))) \\ z_0=\sqrt[4]{18}\cdot(\cos (\frac{\pi}{8})+i\cdot\sin (\frac{\pi}{8}) \\ z_0=\sqrt[4]{18}\cdot e^{i\frac{\pi}{8}} \end{gathered}\begin{gathered} k=1\colon \\ z_1=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{1}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{1}{2}))) \\ z_1=\sqrt[4]{18}\cdot(\cos (\frac{5\pi}{8})+i\cdot\sin (\frac{5\pi}{8})) \\ z_1=\sqrt[4]{18}e^{i\frac{5\pi}{8}} \end{gathered}\begin{gathered} k=2\colon \\ z_2=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{2}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{2}{2}))) \\ z_2=\sqrt[4]{18}\cdot(\cos (\frac{9\pi}{8})+i\cdot\sin (\frac{9\pi}{8})) \\ z_2=\sqrt[4]{18}e^{i\frac{9\pi}{8}} \end{gathered}\begin{gathered} k=3\colon \\ z_3=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{3}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{3}{2}))) \\ z_3=\sqrt[4]{18}\cdot(\cos (\frac{13\pi}{8})+i\cdot\sin (\frac{13\pi}{8})) \\ z_3=\sqrt[4]{18}e^{i\frac{13\pi}{8}} \end{gathered}

Answer:

The four roots in exponential form are

z0 = 18^(1/4)*e^(i*π/8)

z1 = 18^(1/4)*e^(i*5π/8)

z2 = 18^(1/4)*e^(i*9π/8)

z3 = 18^(1/4)*e^(i*13π/8)

5 0
1 year ago
5(3a-1)-2(3a-2)=3(a+2)+v
STatiana [176]

Answer:

<u>The answer is option C. 6a-7</u>

Step-by-step explanation:

Given that

5(3a-1)-2(3a-2)=3(a+2)+v

Solve for v

∴ v = 5(3a-1)-2(3a-2) - 3(a+2)

∴ v = 15a - 5 - 6a + 4 - 3a - 6

∴ v = 15a - 6a - 3a - 5 + 4 - 6

∴ v = 6a - 7

<u>So the answer is option C. 6a-7</u>

5 0
3 years ago
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