All categories

3 years ago

x+2y+2z=3. 2x-2y-z=3. x-2z=-5

Mathematics

1 answer:

d1i1m1o1n [39]3 years ago

8 0

Answer:

x = 1, y = -2, z = 3.

Step-by-step explanation:

x + 2y + 2z = 3 A

2x -2y - z = 3 B

x - 2z = -5 C

If we add equations A and B we eliminate y:

3x + z = 6 E

Multiply E by 2 :

6x + 2z = 12 F

Adding C and F we get:

7x = 7

x = 1.

Plug this into equaton C:

1 - 2z = -5

-2z = -5 - 1 = -6

z = 3.

Now plugging x = 1 and z = 3 in equation A:

1 + 2y + 2(3) = 3

2y = 3 - 1 - 6 = -4

y = -2.

You might be interested in

Solve the inequality 2-x<7

dedylja [7]

Answer:

x > -5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find (f -9)x).<br> f(x) = -3x + 2<br> g(x)= x2

algol13

Answer:

29 and -18

Step-by-step explanation:

f(x)= -3x+12

f(-9)= -3(-9)+2

f(-9)= 27+2

f(-9)= 29

g(x)= x2

g(-9)= -9(2)

g(-9)= -18

7 0

3 years ago

Tickets to a concert cost $70, or $20 with a student discount. Ticket sales must exceed $500,000 for the group to perform. If 20

Angelina_Jolie [31]

Solution:

Let x = the number of $70 seats sold

Let y = the number of $20 seats sold

x+y ≤ 20,000

70x+20y > 500,000

4 0

3 years ago

Read 2 more answers

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

3 years ago

The equation for a circle is x^2 - 8x + y^2 -8 =0. what is the equation of the circle in standard form?

yKpoI14uk [10]

You have to complete the square on this to get it into standard form of a circle. Move the 8 over to the other side because that's part of the radius. Group together the x terms, take half the linear term which is 8, square it and add it in to both sides. Half of 8 is 4, 4 squared is 16, so add in 16 to both sides. I'll show you in a sec. You don't need to do anything to the y squared term. This just means that the center of the circle does not move up or down, only side to side, right or left. Here's your completing the square before we simplify it down to its perfect square binomial. $( x^{2} -8x+16)+ y^{2} =8+16$ . Now break down the parenthesis into the perfect square binomial and do the addition of the right: $(x-4) ^{2} + y^{2}=24$ . This is the standard form of a circle that has a center of (4, 0) and a radius of $\sqrt{24}$

8 0

3 years ago

x+2y+2z=3. 2x-2y-z=3. x-2z=-5​

x+2y+2z=3. 2x-2y-z=3. x-2z=-5