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3 years ago

How many solutions do the systems of equations have? y =-1/3 x + 9 y =- 2/6x

1 answer:

7 0

The answer is: No solution.

Explanation:
The slopes are the same (-1/3 is the same as -2/6 because -2/6 can be simplified to -1/3) but the y intercepts are different.

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Nico is saving money for his college education. He invests some money at 9% and 800 less than that amount at 6%. The investments

vampirchik [111]

Answer:

$1800 at 9% and $1000 at 6%

Step-by-step explanation:

Not needed

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3 years ago

2937 x 28377 ? Plz help

Mice21 [21]

Answer:

83343249

Step-by-step explanation:

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3 years ago

Read 2 more answers

F(x)=12x2-4x-8. G(x)=11x-6

AlekseyPX

Answer: Second option.

Step-by-step explanation:

Given the functions f(x) and g(x):

$f(x)=12x^2-4x-8\\\\g(x)=11x-6$

You need to divide them in order to find $(\frac{f}{g})(x)$ asked in the exercise. Then:

$(\frac{f}{g})(x) =\frac{12x^2-4x-8}{11x-6}$

Now, it is important to remember that, by definition, the division by zero is not defined. Therefore, the denominator of the function cannot be zero.

Let's find the value of "x" for which the denominator of the function would be zero:

1. You need to make the denominator equal to zero:

$11x-6=0$

2. Finally, you must solve for "x":

$11x=6\\\\x=\frac{6}{11}$

Therefore, as you can see, the answer is:

$(\frac{f}{g})(x) =\frac{12x^2-4x-8}{11x-6}$ , $x=\frac{6}{11}$

4 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

2 years ago

What is the value of x

fenix001 [56]

(2x+2) = (3x-52)

2x = 3x-54

-x = -54

x = 54

6 0

3 years ago