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3 years ago

1) Given the equation ax + 3 2 = 6, solve for x.

2 answers:

7 0

$ax+32=6$
subtract 32 from both sides
$ax+32-32=6-32$
simplify
$ax=-26$
divide both sides by a
$\frac{ax}{a}= \frac{-26}{a}$
simplify
$x=- \frac{26}{a}$
Hope this helps

ruslelena [56]3 years ago

4 0

Ax + 32 = 6
- 32 = -32
---------------------
ax = -26
---- -----
a a

1x = -26
-----
a

x = -26
-----
a

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The 21st, 37th and 56th term of an A.P. are consecutive terms of a G.P. Find the common ratio of the G.P.

mash [69]

just use what you know about this stuff

(a+36d)/(a+20d) = (a+55d)/(a+36d)

(a+36d)^2 = (a+55d)(a+20d)

a^2+72ad+1296d^2 = a^2+75ad+1100d^2

3ad = 196d^2

3a = 196d

That is, for any value of n,

a=196n

d=3n

So, there is no unique solution.

If n=1, then a=196 and d=3. The terms are

196+20*3 = 256

196+36*3 = 304

196+55*3 = 361

304/256 = 361/304

You can easily verify that it works for any value of n.

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3 years ago

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Assume z = x + iy, then find a complex number z satisfying the given equation. d. 2z8 – 2z4 + 1 = 0

kodGreya [7K]

Answer: complex equations has n number of solutions, been n the equation degree. In this case:

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i11,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}$

Step-by-step explanation:

I start with a variable substitution:

$Z^{4} = X$

Then:

$2X^{2}-2X+1=0$

Solving the quadratic equation:

$X_{1} =\frac{2+\sqrt{4-4*2*1} }{2*2} \\X_{2} =\frac{2-\sqrt{4-4*2*1} }{2*2}$

$X=\left \{ {{0,5+0,5i} \atop {0,5-0,5i}} \right.$

Replacing for the original variable:

$Z=\sqrt[4]{0,5+0,5i}$

or $Z=\sqrt[4]{0,5-0,5i}$

Remembering that complex numbers can be written as:

$Z=a+ib=|Z|e^{ic}$

Using this:

$Z=\left \{ {{{\frac{\sqrt{2}}{2} e^{i45°} } \atop {{\frac{\sqrt{2}}{2} e^{i-45°} }} \right.$

Solving for the modulus and the angle:

$Z=\left \{ {{\sqrt[4]{\frac{\sqrt{2}}{2} e^{i45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i45}} } \atop {\sqrt[4]{\frac{\sqrt{2}}{2} e^{i-45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i-45}} }} \right.$

The possible angle respond to:

$RAng_{12...n} =\frac{Ang +360*(i-1)}{n}$

Been "RAng" the resultant angle, "Ang" the original angle, "n" the degree of the root and "i" a value between 1 and "n"

In this case n=4 with 2 different angles: Ang = 45º and Ang = 315º

Obtaining 8 different angles, therefore 8 different solutions.

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netineya [11]

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Please help! Find the domain of y = 4 square root 4x + 2

olga55 [171]

Answer:

x ≥ -1/2

Step-by-step explanation:

We know that we cannot graph imaginary numbers. Therefore, our x value has to be greater than or equal to 0:

To find our domain, we need to set the square root equal to zero:

√(4x + 2) = 0

4x + 2 = 0

4x = -2

x = -1/2

We now know that no value below -1/2 can be used or we will get an imaginary number. Therefore, our answer is x ≥ -1/2

Alternatively, we can graph the function and analyze domain:

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Marina CMI [18]

Answer:

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Step-by-step explanation:

First, we need to work out the left-hand side.

find g(2) → g(2) = 2 - 1 which is 1
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Now, we can work out the right hand side.

find f(-1) → f(-1) = -1³ which is -1
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Finally we can work out the full sum: 1 + -2 = -1.

Hope this helps!

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