All categories

3 years ago

Can someone answer these 2 questions for me?

Mathematics

1 answer:

Valentin [98]3 years ago

4 0

For the first question x is equal to 4. You get this by adding 7x+2 and 3x and equal it to 42. And then solve algebraically so 10x+2=42
10x=40 so x equals 4
For the second question x is equal to 6.666 and y is equal to 9
You get this by getting x is to 10 as 4 is to 6 and you get y by 6 is to y as 4 is to 6
Hope this helps

You might be interested in

How to prove tan z is analytic using cauchy-riemann conditions

Basile [38]

A function $f(z)=f(x+iy)=u(x,y)+i v(x,y)$ is analytic if the C-R conditions are satisfied:

$\begin{cases}u_x=v_y\\u_y=-v_x\end{cases}$

We have

$f(z)=\tan z=\tan(x+iy)$

Recall the angle sum formula for tangent:

$\tan(x+iy)=\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}$

Now recall that

$\tan iy=\dfrac{\sin iy}{\cos iy}=\dfrac{-\frac{e^y-e^{-y}}{2i}}{\frac{e^y+e^{-y}}2}=\dfrac{i\sinh y}{\cosh y}=i\tanh y$

So we have

$\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}=\dfrac{\tan x+i\tanh y}{1-i\tan x\tanh y}$
$=\dfrac{(\tan x+i\tanh y)(1+i\tan x\tanh y)}{(1-i\tan x\tanh y)(1+i\tan x\tanh y)}$
$=\dfrac{\tan x+i\tanh y+i\tan^2x-\tan x\tanh^2y}{1+\tan^2x\tanh^2y}$
$=\dfrac{\tan x(1-\tanh^2y)}{1+\tan^2x\tanh^2y}+i\dfrac{\tanh y(1+\tan^2x)}{1+\tan^2x\tanh^2y}$
$=\underbrace{\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}}_{u(x,y)}+i\underbrace{\dfrac{\tanh y\sec^2x}{1+\tan^2x\tanh^2y}}_{v(x,y)}$

We could stop here, but taking derivatives may be messy and would be easier to do if we can write this in terms of sines and cosines.

$u(x,y)=\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}=\dfrac{\frac{\sin x}{\cos x\cosh^2x}}{1+\frac{\sin^2x\sinh^2y}{\cos^2x\cosh^2y}}=\dfrac{\sin x\cos x}{\cos^2x\cosh^2y+\sin^2x\sinh^2y}$
$u(x,y)=\dfrac{\sin2x}{2\cos^2x\cosh^2y+2(1-\cos^2x)(\cosh^2y-1)}=\dfrac{\sin2x}{2\cosh^2y-1+2\cos^2x-1}$
$u(x,y)=\dfrac{\sin2x}{\cos2x+\cosh2y}$

With similar usage of identities, we can find that

$v(x,y)=\dfrac{\sinh2y}{\cos2x+\cosh2y}$

Now we check the C-R conditions.

$u_x=\dfrac{2\cos2x(\cos2x+\cosh2y)-(-2\sin2x)\sin2x}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cos2x\cosh2y}{(\cos2x+\cosh2y)^2}$
$v_y=\dfrac{2\cosh2y(\cos2x+\cosh2y)-(2\sinh2y)\sinh2y}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cosh2y\cos2x}{(\cos2x+\cosh2y)^2}$
$\implies u_x=v_y$

Similarly, you can check that $u_y=-v_x$ , hence the C-R conditions are satisfied, and so $\tan z$ is analytic.

6 0

3 years ago

Mr. Weaver has 36 history students. The ratio of girls to boys is 2:1. How many girls does he have in his class?

mina [271]

Answer:

24 girls

Step-by-step explanation:

24:12

2:1

5 0

3 years ago

Determine whether each statement is true or false. If a statement is true, give a reason or cite an appropriate statement from t

Over [174]

Answer:

1) a. False, adding a multiple of one column to another does not change the value of the determinant.

2) d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.

Step-by-step explanation:

1) If the multiple of one column of a matrix A is added to another to form matrix B then we get: |A| = |B|. Here, the value of the determinant does not change. The correct option is A

a. False, adding a multiple of one column to another does not change the value of the determinant.

2) Two matrices can be column-equivalent when one matrix is changed to the other using a sequence of elementary column operations. Correc option is d.

d. True, column-equivalent matrices are matrices that can be obtained from each other by performing elementary column operations on the other.

3 0

3 years ago

2=x²-10x HOW DO I DO IT???????

almond37 [142]

$2=x^2-10x\\
x^2-10x-2=0\\
x^2-10x+25-27=0\\
(x-5)^2=27\\
x-2=-\sqrt{27} \vee x-2=\sqrt{27}\\
x=2-3\sqrt{3} \vee x=2+3\sqrt{3}$