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2 years ago

PLEASE HELP ASAP- solve the equation : 9x^2+64=0 . i need work to be shown for credit ....

1 answer:

6 0

In this quadratic equation a=9 b=0 c=64, you could use the Quadratic Formula:
x = [-b +-sqrt (b^2 -4*a*c)] / 2*a
x = 0 +- sqrt (0 -4*9*64) / 2*9
x = +- sqrt(-2,304) / 18

x = + 48 i / 18
x = -48 i /18

Where "i" is the square root of minus 1.

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Me. Adams divides 223 markers equally among the 26 students in his class. He puts the extra markers in a box. What is the least

Soloha48 [4]

He puts 15 markers in the box. Hope this helps.

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2 years ago

At the city museum, child admission is $5.20 and adult admission is $9.90 . On Monday, twice as many adult tickets as child tick

DiKsa [7]

To solve this, I would just find what two adult tickets and one child ticket wold be worth.

To find that, just do:

5.20 + 9.90 * 2, which makes 25.

So now you just find how many 25s are in 900, or 36.

So 36 children's tickets were sold that day.

Hope I helped :)

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2 years ago

Solve for x. 12(x+11) = 24

Mars2501 [29]

Answer:

x = -9

Step-by-step explanation:

First, divide both sides by 12 to isolate the expression, x + 11

12(x + 11)/12 = 24/12

x + 11 = 2

Then, subtract 11 from both sides

x + 11 = 2

- 11 - 11

x = -9

4 0

2 years ago

Read 2 more answers

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0

8 months ago

Find the value of x. Round your answer to the nearest tenth

Sati [7]

$x=13.8$