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kupik [55]
3 years ago
13

What is the factored expression of 63x^2-3x-6

Mathematics
1 answer:
Murljashka [212]3 years ago
6 0

Answer:

3(3x-1)(7x +2)

Step-by-step explanation:

63x^2-3x-6

Suppose a generic quadratic equation

ax^2 + bx + c

To factor this equation, I need to find its roots. Then we use the quadratic formula of:

\frac{-b + \sqrt{b^2-4ac}}{2a}

and

\frac{-b + \sqrt{b^2-4ac}}{2a}

So, for the equation 63x ^ 2-3x-6 we have:

\frac{3 + \sqrt{(-3)^2-4(63)(-6)}}{2(63)} = \frac{1}{3}

and

\frac{3 - \sqrt{(-3)^2-4(63)(-6)}}{2(63)} = \frac{-2}{7}

So:

(x-\frac{1}{3}) = 0\\\\(3x-1) = 0

and

(x - (-\frac{2}{7})) = 0\\\\(x+ \frac{2}{7}) = 0\\\\(7x +2) = 0

Finally the polynomial is:

(3x-1)(7x +2)

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astra-53 [7]

Answer:

See attached. (Both inequalities graph to the same region.)

Step-by-step explanation:

To graph an inequality, you first replace the inequality symbol by an equal sign. Then you graph the line (or curve) described by the equation. If the inequality symbol includes the "or equal to" case, then the line (or curve) is solid and is part of the solution set. If not, then the line (or curve) is dashed and is <em>not part of the solution set</em>.

Here, the line x+y=5, will bound the solution set, but is not part of it.

Note that the line -x -y = -5 (from the second inequality) is the same ilne.

_____

After you have graphed the boundary of the solution region, you need to determine which side of the boundary has the solutions. I find it easy to do this by locating an x- or y-term with a positive coefficient and seeing which side of the inequality symbol that lies on. Here, both the x-term and the y-term lie on the "is greater than" side of the symbol, so the half-plane is shaded above and to the right of the line, where x- and y-values are greater than they are on the line.

In the second inequality, you can multiply the whole thing by -1 to get

... x + y > 5 . . . . same as the first inequality

or you can add (x+y) to both sides to get

... 0 < -5 +x +y

Once again, the x- and y-values that satisfy this are greater than those that are on the boundary line.

_____

The graph seems to show only one equation. Both are there, but the dashed blue line sits on top of the dashed red line, so you can't see it. The purple shading indicates both red and blue shading in the same area.

===== =====

Of course you graph the boundary line the way you would any linear equation. In this form, it is not too difficult to see that both the x- and y-intercepts are 5. (When x=0, y=5; and when y=0, x=5 for the boundary line's equation x+y=5.) Draw the dotted line through these intercept points, add shading above, and you have your graph.

6 0
3 years ago
Find the value of g(7) for the function below.
Yanka [14]

Answer: The correct answer is (A) - 60/7 :)

Step-by-step explanation: The answer is A because all of the other ones make absolutely no sense one bit at all...!

5 0
3 years ago
The Kendal Company manufactures automobile engine parts. Running the manufacturing operations requires 12 watts (units of electr
Monica [59]

24 because when you plug in 0 for both x and y:  

0=12(0)+24  

12*0=0, leaving you with 24.

4 0
3 years ago
A rhombus has four 6-inch sides and two 120-degree angles. From one of the vertices of the obtuse angles, the two latitudes are
nikitadnepr [17]

Answer:

Area(A)=Area(C)= 9 in^{2}

Area(B)=13.2 in^{2}

Step-by-step explanation:

We begin with finding the angles a and b that from the drawing attached you can see that a=b.

Now, the sum of the internal angles of a rhomboid is equal to 360 degrees, with that we have:

120+120+a+b=360

240+2a=360

2a=120

a=60=b

Next, in the image you can see that the lines coming from the angle at the top 120 degrees vertex, divide the opposite sides by half, thus making two triangles with one side of 6 in and another of 3 in.

We can say from the drawing as well:

Area(A)+Area(B)+Area(C)=Area(rhomboid)

But, we can also say that Area(A)=Area(C)

So, starting with Area(A)

Area(A)=Area(triangle)=\frac{b*h}{2}=\frac{6*3}{2}=9 in^{2}

We can then calculate the area B, a rhomboid, or better, take the Total area of the figure and subtract the area of the two triangles.

Area(B)=Area(rhomboid)-Area(A)-Area(C)

Area(rhomboid)=b*h where b=6in and h is the perpendicular distance from the base to the top.

h=[tex]6*cos(30)=5.20in   The 30 degrees come from: 120-30-60=30, since the latitudes split the 120 angle in two equal parts and one that is the half of the obtuse angle.

Area(rhomboid)=5.20*6=31.2 in^{2}

Area(B)=Area(rhomboid)-Area(A)-Area(C)=31.2 in^{2}-9 in^{2}-9 in^{2}=13.2 in^{2}

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%2F7%7D%7B7%5Csqrt%7B10%7D%2F2%20%7D" id="TexFormula1" title="\frac{3/7}{7\sqrt{10}
VashaNatasha [74]

Multiply the numerator and denominator by 7×2 = 14 to eliminate the denominators of those fractions:

\dfrac{\dfrac37}{\dfrac{7\sqrt{10}}2}\times\dfrac{14}{14}=\dfrac{3\times2}{7\sqrt{10}\times7}=\dfrac6{49\sqrt{10}}

Rationalize the denominator by multiplying both numerator and denominator by √10:

\dfrac6{49\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{6\sqrt{10}}{49(\sqrt{10})^2}=\dfrac{6\sqrt{10}}{49\times10}=\dfrac{6\sqrt{10}}{490}

Lastly, cancel the common factor of 2 in both the numerator and denominator (which comes from 6 = 2×3 and 490 = 2×245):

\dfrac{6\sqrt{10}}{490}=\dfrac{3\sqrt{10}}{245}}

8 0
3 years ago
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