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3 years ago

Helpppppppppppppppppp

Mathematics

1 answer:

UNO [17]3 years ago

8 0

Answer:

2 solutions

Step-by-step explanation:

Given the expression

2m/2m+3 - 2m/2m-3 = 1

Find the LCM of the expression at the left hand side:

2m(2m-3)-2m(2m+3)/(2m+3)(2m-3) = 1

open the bracket

4m²-6m-4m²-6m/(4m²-9) = 1

Cross multiply

4m²-6m-4m²-6m = 4m² - 9

-12m = 4m² - 9

4m² - 9+12m = 0

4m² +12m-9 = 0

Since the resulting equation is a quadratic equation, it will have 2 solutions since the degree of the equation is 2

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I will mark who ever gets it correct the brainiest. Help Asap. See image below.

marysya [2.9K]

The transformation from the first equation to the second one can be found by finding
a
a
,
h
h
, and
k
k
for each equation.
y
=
a
|
x
−
h
|
+
k
y
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a
|
x
-
h
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+
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x
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y
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x
|
Factor a
1
1
out of the absolute value to make the coefficient of
x
x
equal to
1
1
.
y
=
|
x
|
−
4
y
=
|
x
|
-
4
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a
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h
h
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k
k
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y
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x
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−
4
y
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x
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-
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a
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=
1
h
=
0
h
=
0
k
=
−
4
k
=
-
4
The horizontal shift depends on the value of
h
h
. When
h
>
0
h
>
0
, the horizontal shift is described as:
g
(
x
)
=
f
(
x
+
h
)
g
(
x
)
=
f
(
x
+
h
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g
(
x
)
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(
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−
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-
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Horizontal Shift: None

4 0

3 years ago

Given the following statements, which answer uses deductive reasoning to draw the conclusion?

MatroZZZ [7]

Determine whether the stated conclusion is valid based on the given information. If not, write invalid. Explain your reasoning.
3. Given: If a number is divisible by 4, then the number is divisible by 2. 12 is divisible by four.
Conclusion: 12 is divisible by 2.
SOLUTION:
If p → q is a true statement and p is true, then q. Here, a number divisible 4 is divisible by 2 also and 12 is divisible by 4. So, 12 is divisible by 2 is a valid statement by the Law of Detachment.
ANSWER:
valid; Law of Detachment
4. Given: If Elan stays up late, he will be tired the next day. Elan is tired.
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If p is true, then q is true, but q being true does not necessarily mean that p is true. Elan could be tired because he worked out. So, the statement is invalid.
ANSWER:
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8 0

3 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:

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