All categories

2 years ago

Richard and Stephen win some money and share it in the ratio 1:3. Richard gets £14. How much did Stephen get?

Mathematics

1 answer:

LenaWriter [7]2 years ago

5 0

Well depending on who has what ratio Stephen could have £42 or £4.66

You might be interested in

Mary’s cookie jar opening has a circumference of 24 inches.

HACTEHA [7]

Answer:

75.40 inches

Step-by-step explanation:

If it is its circumference, then the answer is self-evident, 24 inches. If it is its diameter, then we must do a little calculation, multiplying the diameter by Pi to reach 75.40 inches.

5 0

2 years ago

Read 2 more answers

The diagonals of a rumbus are 21m and 32m what is area of the rombus

Liula [17]

Consider this option:
1. formula is: S=0.5*d₁*d₂, where d₁;d₂ - the diagonals of the rhombus.
2. substituting the values into the formula: S=0.5*21*32=336 m².

answer: 336 m².

5 0

2 years ago

Read 2 more answers

What us the problem to 5^4×3^2

torisob [31]

Answer:

The answer is 5625

Step-by-step explanation:

Lets break up the problem

So you have two things: 5^4 and 3^2

Solve seperately

Gives you 625 and 9

Then multiply both together

625*9 = 5625

4 0

3 years ago

Read 2 more answers

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:

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

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

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

A line passing (a,3) and (5,5) is perpendicular to a line passing through (0,0) and (1,a) Find the value of a.

vodka [1.7K]

Answer:

a = -5

Step-by-step explanation:

(5 - 3)/(5 - a)

2/(5 - a)

perpendicular of 2/(5 - a) = -(5 - a)/2

(a - 0)/(1 - 0) = -(5 - a)/2

a/1 = -(5 - a)/2

2a = -5 + a

a = -5

6 0

3 years ago