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yanalaym [24]
3 years ago
8

Which of the following is the solution set of x2 + 8x + 12 = 0? (-2.-6 2.6

Mathematics
1 answer:
algol [13]3 years ago
6 0

To solve this polynomial equation, we first factor the left side.

To factor x² + 8x + 12 = 0, look for factors of 12 that add to 8.

These factors are 6 and 2 so we have (x + 6)(x + 2) = 0.

So either x + 6 = 0 or x + 2 = 0 and solving for

x in each equation, our answer is {-6, -2}.

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What fractions are in between -1/3 any -2/3
UNO [17]

To find a fraction between two fractions, all we need to do is make the sum of the numerators be the new numerator, and the sum of the denominators be the new denominator.(x) So, for example, a fraction between 7/13 and 6/11 is (7 + 6)/(13+ 11) =13/24.(x)

7/13 = .5384615(x)

6/11 = .545454(x)

13/24 = .541666�

 

Given that a/b < c/d, why is it true that a/b < (a+c)/(b+d)< c/d?

6 0
3 years ago
Find the values of x and y.
kupik [55]

Answer:

x=6, y=75

Step-by-step explanation:

Since the diameter of the circle is a perpendicular bisector to the chord, x and 6 must be equal and therefore x=6. It also bisects the arc, meaning that y=75. Hope this helps!

4 0
4 years ago
Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).
Elan Coil [88]

Answer: \int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. <u>Normal</u> <u>vector</u> is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = \left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]

n = 36i + 0j + 45k - (0k + 0i - 20j)

n = 36i + 20j + 45k

Equation of a plane is generally given by:

a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0

Then, replacing with point P and normal vector n:

36(x-5) + 20(y-0) + 45(z-0) = 0

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

z = \frac{180-36x-20y}{45}

When z = 0:

y = 9 + \frac{-9x}{5}

When z=0 and y=0:

x = 5

Then, triple integral is:

\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx

Calculating:

\int\limits^5_0 {\int\limits {\int\ {xyz}  \, dy } \, dx

\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )}  \, dy } \, dx

\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}}  \, dy } \, dx

\frac{1}{45} \int\limits^5_0  {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx

\frac{1}{45} \int\limits^5_0  {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4}  } \, dx

\frac{1}{45} [30375-60750+118462.5-39150]

\int\limits^5_0 {\int\limits {\int\ {xyz}  \, dy } \, dx = 1087.5

<u>The volume of the tetrahedon is 1087.5 cubic units.</u>

3 0
3 years ago
Find the value if x that will make A||B
enot [183]

Answer:

x=11

Step-by-step explanation:

The angles given are alternate exterior angles and alternate exterior angles are equal if the lines are parallel

5x+9 = 6x-2

Subtract 5x from each side

5x+9-5x = 6x-2-5x

9 = x-2

Add 2 to each side

9+2 =x-2+2

11=x

7 0
3 years ago
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Write an expression for b divided by 6
Nonamiya [84]
B divided by 6

b / 6...." / " means divide
7 0
3 years ago
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