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3 years ago

How do you solve 3x^6- 39x^4+ 108x^2= 0

1 answer:

6 0

You can first take out an x^2 from each number , 3x^6 - 39x^4 + 108x^2 = x^2 ( 3x^4 - 39x^2 + 108) Then just factor and solve for the zero's of the equation

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The roots of the function f(x) = x2 – 2x – 3 are shown. what is the missing number? x = –1 and x =

andrey2020 [161]

we have

$f(x)=x^{2}-2x-3$

Find the roots

Equate the function to zero

$x^{2}-2x-3=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=3$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=3+1$

$x^{2}-2x+1=4$

Rewrite as perfect squares

$(x-1)^{2}=4$

Square root both sides

$(x-1)=(+/-)2$

$x=1(+/-)2$

$x=1+2=3$

$x=1-2=-1$

therefore

<u>the answer is</u>

$x=3$

5 0

3 years ago

Read 2 more answers

Use the distributive property to write -2(x + 5) as an equivalent expression

mamaluj [8]

Answer:

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Step-by-step explanation: oh night longg

6 0

3 years ago

The curved part of this figures is a semicircle.

vovangra [49]

First we join the two endpoints of the semicircle and that will be the diameter.

And to find the length of the diameter, we have to use distance formula, with one endpoint (3,2) and the other is (-4,-2) .

SO we get

$Diameter = \sqrt{ {-4-3)^2 +(-2-2)^2 } = \sqrt{49+16} = \sqrt 65$

Radius is half of diameter, so the radius is

$Radius= \frac{ \sqrt{65}}{2}$

Formula of area of circle is

$Area = \pi r^2$

So the area of semicircle is

$=(1/2) \pi ( \frac{ \sqrt{65}}{2})^2 = 8.125 \pi \square \units$

And the other figure is a triangle, with

$Base = 3-(-4)=3+4=7 \\ height = 2-(-2) = 2+2 =4$

$Area = (1/2)*4*7= 14 square units$

Therefore, total area is the sum of area of semicircle and area of triangle,

And we will get

$Total \ area= 14 + 8.125 \pi \ units^2$

Correct option is the third option .

3 0

3 years ago

Read 2 more answers

if a guy has $145 .00 in his hand, and he wants to give some to 2 friends how much is left in his hand

Mariulka [41]

$145.00/2=
$72.5
I hopes this helped

7 0

3 years ago

Read 2 more answers

Match the numerical expressions to their simplest forms.

Aloiza [94]

Answer:

$(a^6b^1^2)^\frac{1}{3}$ = $a^2b^4$

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$ = $a^3b^2$

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$ = $a^2b$

$(\frac{a^3}{ab^-^6})^\frac{1}{2}$ = $ab^3$

Step-by-step explanation:

Simplify each of the expressions:

$(a^6b^1^2)^\frac{1}{3}$

Distribute the exponent. Multiply the exponent of the term outside of the parenthesis by the exponents of the variable.

$(a^6b^1^2)^\frac{1}{3}$

$a^6^*^\frac{1}{3}b^1^2^*^\frac{1}{3}$

Simplify,

$a^2b^4$

Use a similar technique to solve this problem. Remember, a fractional exponent is the same as a radical, if the denominator is (2), then the operation is taking the square root of the number.

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$

Rewrite as square roots:

$\frac{\sqrt{a^5b^3}}{\sqrt{(ab)}^-^1}$

A negative exponent indicates one needs to take the reciprocal of the number. Apply this here:

$\frac{\sqrt{a^5b^3}}{\frac{1}{\sqrt{ab}}}$

Simplify,

$\sqrt{a^5b^3}*\sqrt{ab}$

Since both numbers are under a radical, one can rewrite them such that they are under the same radical,

$\sqrt{a^5b^3*ab}$

Simplify,

$\sqrt{a^6b^4}$

Since this operation is taking the square root, divide the exponents in half to do this operation:

$a^3b^2$

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$

Simplify, to simplify the expression in the numerator and the denominator, the base must be the same. Remember, the base is the number that is being raised to the exponent. One subtracts the exponent of the number in the denominator from the exponent of the like base in the numerator. This only works if all terms in both the numerator and the denominator have the operation of multiplication between them:

$(\frac{a^8}{b^-^4})^\frac{1}{4}$

Bring the negative exponent to the numerator. Change the sign of the exponent and rewrite it in the numerator,

$(a^8b^4)^\frac{1}{4}$

This expression to the power of the one forth. This is the same as taking the quartic root of the expression. Rewrite the expression with such,

$\sqrt[4]{a^8b^4}$

SImplify, divide the exponents by (4) to simulate taking the quartic root,

$a^2b$

$(\frac{a^3}{ab^-^6})^\frac{1}{2}$

Using all of the rules mentioned above, simplify the fraction. The only operation happening between the numbers in both the numerator and the denominator is multiplication. Therefore, one can subtract the exponents of the terms with the like base. The term in the denomaintor can be rewritten in the numerator with its exponent times negative (1).

$(a^3^-^1b^(^-^6^*^(^-^1^)^))^\frac{1}{2}$

$(a^2b^6)^\frac{1}{2}$

Rewrite to the half-power as a square root,

$\sqrt{a^2b^6}$

Simplify, divide all of the exponents by (2),

$ab^3$

7 0

3 years ago