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denis-greek [22]
3 years ago
11

I have no clue how to do this someone help

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
3 0
So here’s a example and formula

You might be interested in
Solve 5x(x-6)(4x+9)=0
xxTIMURxx [149]
Greetings and Happy Holidays!

Solve 
for x.
5x(x-6)(4x+9)=0<span>

Distribute the parenthesis:
</span>5x(x-6)(4x+9)=0

(5x^2-30x)(4x+9)=0

(20x^3-120x^2)+(45x^2-270x)=0

Combine like terms:
20x^3-75x^2-270x=0

Factor out GCF (5x):
5x(4x^2-15x-54)=0

Factor the Complex Trinomial:
5x(4x^2+9x-24x-54)=0

5x(x(4x+9)-6(4x-9))=0

5x((4x+9)(x-6))=0

Set Factors to Equal 0:
5x=0

x=0

or

4x+9=0

4x=-9

x=\frac{-9}{4}

or

x-6=0

x=6

The Answers Are:
\boxed{x=0} \\ \boxed{x= \frac{-9}{4}} \\ \boxed{x=6}

I hope this helped!
-Benjamin

8 0
3 years ago
X^4-64=0<br>The examples we did in class had solutions like "x=+-2" and "x=+-i(square root)3.​
hammer [34]

Answer:

Step-by-step explanation:

This isn't the same thing although you will get some factors that have i in them

(x^4 - 64) factors using the difference of squares.

(x^2 - 8)(x^2 + 8) Both of these factor using the difference of squares.

(x^2 - 8): factors into (x + sqrt(8) )(x - sqrt(8) )

(x^2 - 8):  factors further (x + 2*sqrt2)(x - 2sqrt(2)

(x ^2 + 8) : (x + sqrt(8)i ) (x - sqrt(8)i )

(x^2 + 8) : (x + 2sqrt(2)i) (x - 2sqrt(2)i)

Final answer

(x + 2*sqrt2)(x - 2sqrt(2)(x + 2*sqrt2)(x - 2sqrt(2)

3 0
3 years ago
Write the equation in standard form. Identity the center and radius.<br> x² + y2 + 8x-4y-7=0
lorasvet [3.4K]

Answer:

equation;

(x + 4) {}^{2}  + (y - 2) {}^{2}  = 27

Center (-4,2)

Radius is

3 \sqrt{3}

Step-by-step explanation:

Since the x^2 and y^2 have the same coeffiecent this will be a circle in a form of

(x - h) {}^{2}  + (y - k)  {}^{2}  =  {r}^{2}

Where (h,k) is center

r is the radius

So first we group like Terms together

{x}^{2}  + 8x  +  {y}^{2}  - 4y - 7 = 0

Add 7 to both sides

{x}^{2}  + 8x +  {y}^{2}  - 4y = 7

( {x}^{2}  + 8x) +(  {y}^{2}  - 4y) = 7

Since the orginal form of the equation of the circle has a perfect square we need to complete the square for each problem

(\frac{8}{2} ) {}^{2}  = 16

and

( - \frac{4}{2} ) {}^{2}  = 4

so we have

{x}^{2}  + 8x + 16 +  {y}^{2}  - 4y + 4y = 7 + 16 + 4

{x}^{2}  + 8x + 16  +  {y}^{2}  - 4y + 4y = 27

(x + 4) {}^{2}  +  (y  - 2) {}^{2}  = 27

To find our center, h is -4 and k is 2

so the center is (-4,2)

The radius is

\sqrt{27}  = 3 \sqrt{3}

So the radius is 3 times sqr root of 3.

8 0
2 years ago
MATH HELP PLEASE!!! USA TESTPREP!!
MaRussiya [10]

Answer:

B.) -3

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
How can we eliminate an imaginary from the denominator? Simplify 2/(3+5i) to demonstrate.
alexira [117]
Multiply by the conjugate and simplify
3/17 - 5i/17

Hope this helps! :)
7 0
3 years ago
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