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AleksAgata [21]
3 years ago
5

Solve 1 over 36 = 6x^−3. plz help

Mathematics
2 answers:
nadya68 [22]3 years ago
5 0
-----------------------
1/36 = 6x⁻³

1/36 = 6 x 1/x^3

1/36 = 6/x^3

1/36x^3 = 6

x^3/36 = 6

x^3 = 6 x 36

x^3 = 216

x = ∛216

x = 6
-----------------------
telo118 [61]3 years ago
4 0

Answer: x = 1

Step-by-step explanation:

Sorry for the delay

Let's solve your equation step-by-step.

1

36

=6x−3

Step 1: Flip the equation.

6x−3=

1

36

Step 2: Solve Exponent.

6x−3=

1

36

log(6x−3)=log(

1

36

)(Take log of both sides)

(x−3)*(log(6))=log(

1

36

)

x−3=

log(

1

36

)

log(6)

x−3=−2

x−3+3=−2+3(Add 3 to both sides)

x=1

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Which is the completely factored form of 3xsquared<br> -12x-15?
tekilochka [14]

Answer:

3(x - 5)(x + 1).

Step-by-step explanation:

3x^2 - 12x - 15

Dividing through by 3:

= 3(x^2 - 4x - 5)

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3 years ago
What is a equivalent fraction with 11/12 and the denominator 24
oksian1 [2.3K]

Answer:

22/24

Step-by-step explanation:

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To do that, multiply the numerator (11) by 2, (because the denominator of the second fraction is 2 times the denominator of the first fraction).

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3 years ago
What is 1/5+1/5???????
KatRina [158]

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Step-by-step explanation:

4 0
4 years ago
Read 2 more answers
ANSWER ASAP PLEASE &amp; SHOW UR WORK!!!
Lina20 [59]

Answer:

x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}

Step-by-step explanation:

<u>Biquadratic Equations</u>

Solve:

x^4 + 3x^2 - 18 = 0

The biquadratic equations are equations of degree 4 without the terms of degree 1 and 3.

Solving such equations requires to express the equation as a second-degree equation with x^2 as the variable.

Rewriting the equation:

(x^2)^2 + 3(x^2) - 18 = 0

The quadratic equation can be factored as:

(x^2+6)(x^2-3)=0

It leads to two equations:

x^2+6=0

x^2-3=0

The first equation has imaginary roots. Solving for x:

x^2=-6

x=\pm\sqrt{-6}

x_1=\mathbf{i}\sqrt{6}

x_2=-\mathbf{i}\sqrt{6}

Where

\mathbf{i}=\sqrt{-1}

The second equation has two real roots:

x^2-3=0

x^2=3

x=\pm\sqrt{3}

x_3=\sqrt{3}

x_4=-\sqrt{3}

The roots are:

\mathbf{x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}}

3 0
3 years ago
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