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2 years ago

What is the square root of 1/25

Mathematics

1 answer:

wolverine [178]2 years ago

6 0

First simplify the square root of 1/25

-√1/25

Rewrite using the quote by rule for radicals.

-√1/√25

Any root of 1 is 1

-1/√25

Simplify the denominator. (find the square root of 25)

-1/5

The result can be shown in both exact and decimal forms.
Exact form: -1/5
Decimal form: -0.2

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How can Ari simplify the following expression? StartFraction 5 Over a minus 3 EndFraction minus 4 divided by 2 + StartFraction 1

kirill [66]

Answer:

The simplification for the expression is given as =( 7 + 2(a-3))/(a-3)

Step-by-step explanation:

To simplify the expression we will first convert the words to values in numbers and alphabets.

StartFraction 5 Over a minus 3 EndFraction minus 4 divided by 2 + StartFraction 1 Over a minus 3 EndFraction

= 5/(a-3) -4/2 + 2/(a-3)

Having done that, let's move on and simplify the expression.

5/(a-3) -4/2 + 2/(a-3)

= 5/(a-3) -2+ 2/(a-3)

= 5/(a-3) + 2/(a-3) -2

= 7/(a-3) -2

=( 7 + 2(a-3))/(a-3)

7 0

2 years ago

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.

3 0

2 years ago

Plz with steps .. it's very hard can anyone plz

liubo4ka [24]

Answer:

Step-by-step explanation:

$\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a} \qquad f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a} \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} }{\dfrac{1}{2\sqrt{x}} }\\\\$

$\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}} =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\$

7 0

2 years ago

Kiana is saving for a new computer that cost $499. So far, she has earned 68 percent of the cost of the computer. How much more

marysya [2.9K]

Answer:

159.68

Step-by-step explanation:

68% of 499 is 339.32

499 -339.32=159.68

hope this helps

6 0

2 years ago

50 Points: Name the similar triangles & Prove the similarity with details

Leno4ka [110]

Let's consider the facts at hand:

By Vertical Angle Theorem ⇒ ∠BCE = ∠DCF
∠BEC = ∠DFC
Sides BE = DF

Based on the diagram, triangles BCE and triangles DCF are similar

⇒ based on the Angle-Angle theorem

⇒ since ∠BCE = ∠DCF and ∠BEC = ∠DFC

⇒ the two triangles are similar

Hope that helps!

Definitions of Theorem I used:

Vertical Angle Theorem: opposite angles of two intersecting lines must be equal
Angle-Angle Theorem: if two angles of both triangles are equal, then the given triangles must be similar

8 0

1 year ago