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4 years ago

Solve the equation for x, where x is a real number (5 points): 5x^2 + 11x - 12 = -10

Mathematics

2 answers:

jenyasd209 [6]4 years ago

5 0

Add 10 and use the quadratic formula.
5x^2 +11x -2 = 0
x = (-11 ±√(11^2 -4(5)(-2)))/(2(5))
x = (-11 ±√161)/10 ≈ {-2.368858, 0.168858}

_____
The quadratic formula tells you that for $ax^{2}+bx+c=0$ the solutions are
$x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

AVprozaik [17]4 years ago

3 0

Add 10 to both sides so that the equation becomes 5x^2 + 11x - 2 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
                     x = [ -b ± √(b^2 - 4ac) ] / (2a)
                     x = [ -11 ± √(11^2 - 4(5)(-2)) ] / ( 2(5) )
                     x = [ -11 ± √(121 - (-40) ) ] / ( 10 )
                     x = [ -11 ± √(161) ] / ( 10)
                     x = [ -11 ± sqrt(161) ] / ( 10 )
                     x = -11/10 ± sqrt(161)/10
The answers are -11/10 + sqrt(161)/10 and -11/10 - sqrt(161)/10.

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Let $A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$. There exists a point $X$ and a constant $k$ such that for any point $P$,

goldfiish [28.3K]

Answer:

k=32

Step-by-step explanation:

Given the points:

$A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$.$

The first step is to find the Centroid of the triangle.

Centroid, X

$=\left(\dfrac{x_1+x_2+x_3}{3} ,\dfrac{y_1+y_2+y_3}{3} \right)\\=\left(\dfrac{4+6+(-1)}{3} ,\dfrac{-1+2+2}{3} \right)\\=\left(\dfrac{9}{3} ,\dfrac{3}{3} \right)=(3,1)$

Next, let P be a point (x,y)

Using the distance formula, $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$PA^2=(x-4)^2+(y-(-1))^2\\PB^2=(x-6)^2+(y-2)^2\\PC^2=(x-(-1))^2+(y-2)^2\\PX^2=(x-3)^2+(y-1)^2\\$

On Substitution into: $PA^2 + PB^2 + PC^2 = 3PX^2 + k$

$(x-4)^2+(y-(-1))^2+(x-6)^2+(y-2)^2+(x-(-1))^2+(y-2)^2=3[(x-3)^2+(y-1)^2]+k$

Let us simplify the LHS first

$\\LHS: x^2-8x+16+y^2+2y+1+x^2-12x+36+y^2\\-4y+4+x^2+2x+1+y^2-4y+4\\=3x^2-18x+3y^2-6y+62$

Also, the Right Hand Side

$RHS:3[(x-3)^2+(y-1)^2]+k\\=3[x^2-6x+9+y^2-2y+1]+k\\=3x^2-18x+27+3y^2-6y+3+k\\=3x^2+3y^2-18x-6y+30+k$

Therefore:

$3x^2-18x+3y^2-6y+62=3x^2+3y^2-18x-6y+30+k\\k=3x^2-18x+3y^2-6y+62-3x^2-3y^2+18x+6y-30\\k=3x^2-3x^2+3y^2-3y^2-18x+18x+62-30\\k=32$

7 0

3 years ago

If a vehicle travels 25x miles in 2 hours, what is the vehicle's average speed, in miles per hour.

il63 [147K]

Given that a vehicle travels 25 miles in 2 hours, determine the average speed of the vehicle in miles per hour.

Work:

In order to find the average mph of the vehicle, you have to divide the distance the vehicle travels in a certain period of time by that amount of time.

So, in this case, the vehicle traveled 25 miles in 2 hours, so you would divide 25 by 2.

25/2 = 12.5

That means, the vehicle is traveling at a speed of 12.5 mph.

Thus, the average speed of the vehicle is 12.5 miles per hour (mph).

3 0

4 years ago

The game of clue involves 6 suspects, 6 weapons, and 9 rooms. one of each is randomly chosen and the object of the game is to gu

mina [271]

Part A:

Given that the game of clue involves 6 suspects, 6 weapons, and 9 rooms.

The number of ways that one of each is randomly chosen is given by:

$^6C_1\times{ ^6C_1}\times{ ^9C_1}=6\times6\times9=324$

Therefore, the number of solutions possible is 324.

Part B:

Given that a players is randomly given three of the remaining cards, let s, w, and r be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player.

The number of suspects, weapons, and rooms remaining respectively after the player observes his or her three cards are: 6 - s, 6 - w, and 9 - r.

Let x denote the number of solutions that are possible after that player observes his or her three cards, then:

$x={ ^{6-s}C_1}\times{ ^{6-w}C_1}\times{ ^{9-r}C_1}=(6-s)(6-w)(9-r)$

Therefore, x in terms of s, w, and r is given by x = (6 - s)(6 - w)(9 - r).

Part C:

The expected value E(x) of a data set $x_i$ with probabilities $p(x_i)$ is given by $E(x)=\Sigma xp(x)$

There are $^{3+3-1}C_{3-1}={ ^5C_2}=10$ possible combinations s, w and r. They are (3, 0, 0), (0, 3, 0), (0, 0, 3), (2, 1, 0), (0, 2, 1), (1, 0, 2), (2, 0, 1), (1, 2, 0), (0, 1, 2), (1, 1, 1)

Thus the expected value is given by

$E(x)=3\cdot6\cdot9p(3, 0, 0)+6\cdot3\cdot9p(0, 3, 0)+6\cdot6\cdot6p(0, 0, 3) \\ 4\cdot5\cdot9p(2, 1, 0)+6\cdot4\cdot8p(0, 2, 1)+5\cdot6\cdot7p(1, 0, 2)+4\cdot6\cdot8p(2, 0, 1) \\ +5\cdot4\cdot9p(1, 2, 0)+6\cdot5\cdot7p(0, 1, 2)+5\cdot5\cdot8(1, 1, 1) \\ \\ = \frac{1}{ ^{21}C_3} (162\cdot{ ^6C_3}\cdot{ ^6C_0}\cdot{ ^9C_0}+162\cdot{ ^6C_0}\cdot{ ^6C_3}\cdot{ ^9C_0}+216\cdot{ ^6C_0}\cdot{ ^6C_0}\cdot{ ^9C_3} \\ \\ +180\cdot{ ^6C_2}\cdot{ ^6C_1}\cdot{ ^9C_0}+192\cdot{ ^6C_0}\cdot{ ^6C_2}\cdot{ ^9C_1}$

$+210\cdot{ ^6C_1}\cdot{ ^6C_0}\cdot{ ^9C_2}+192\cdot{ ^6C_2}\cdot{ ^6C_0}\cdot{ ^9C_1}+180\cdot{ ^6C_1}\cdot{ ^6C_2}\cdot{ ^9C_0} \\ \\ +210\cdot{ ^6C_0}\cdot{ ^6C_1}\cdot{ ^9C_2}+200\cdot{ ^6C_1}\cdot{ ^6C_1}\cdot{ ^9C_1} \\ \\ =\frac{1}{1,330}(324\cdot20+216\cdot84+360\cdot90+384\cdot135+420\cdot216+200\cdot324) \\ \\ =\frac{1}{1,330}(6,480+18,144+32,400+51,840+90,720+64,800) \\ \\ =\frac{1}{1,330}(264,384) \\ \\ =\bold{198.78}$

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3 years ago

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Zigmanuir [339]

Answer:

9. C. division

10. D. addition

Step-by-step explanation:

When solving for X the goal is to get it all by itself on one side of the equation and to do that you have to undo the operations already in effect on it. In number 9 the x is multiplied by 3 so to undo that you would need to divide by 3 and that would cancel out leaving the x by itself. Then you would need to divide by 3 on the other side of the equation also to keep it balenced. In number 10 the X is connected to the 10 through subtraction so to cancel that out you would subtract 10 from both sides to get the X alone.

I hope this helps and please don't hesitate to ask if there is anything still unclear!

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