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damaskus [11]
3 years ago
15

Find the difference. Write your answer in simplest form. 4 - 1 1/4

Mathematics
1 answer:
4vir4ik [10]3 years ago
3 0
5/4 should be correct
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The answer is k = 2
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Which statements must be true about the image of ΔMNP after a reflection across Line E G? Select three options. The image will b
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Answer:

The image will be congruent to ΔMNP.

Line EG will be perpendicular to the line segments connecting the corresponding vertices.

The line segments connecting corresponding vertices will all be parallel to each other.

Step-by-step explanation:

The reflected image will be the mirror image of ΔMNP, so it will be congruent.  The lines connecting each vertex to the corresponding vertex will be parallel, and all three lines are perpendicular to EG.

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Photo attached. Thanks in advance!
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The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.
andrezito [222]

Answer:

Given the equation: 3x^2+10x+c =0

A quadratic equation is in the form: ax^2+bx+c = 0 where a, b ,c are the coefficient and a≠0 then the solution is given by :

x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a} ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}

Simplify:

x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}

Also, it is given that the difference of two roots of the given equation is 4\frac{2}{3} = \frac{14}{3}

i.e,

x_1 -x_2 = \frac{14}{3}

Here,

x_1 = \frac{-10 + \sqrt{100- 12c}}{6} ,     ......[2]

x_2= \frac{-10 - \sqrt{100- 12c}}{6}       .....[3]

then;

\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}

simplify:

\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}

or

\sqrt{100- 12c} = 14

Squaring both sides we get;

100-12c = 196

Subtract 100 from both sides, we get

100-12c -100= 196-100

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve x_1 , x_2

x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}

or

x_1 = \frac{-10 + \sqrt{100- 96}}{6} or

x_1 = \frac{-10 + \sqrt{4}}{6}

Simplify:

x_1 = \frac{-4}{3}

Now, to solve for x_2 ;

x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}

or

x_2 = \frac{-10 - \sqrt{100- 96}}{6} or

x_2 = \frac{-10 - \sqrt{4}}{6}

Simplify:

x_2 = -2

therefore, the solution for the given equation is: -\frac{4}{3} and -2.


3 0
3 years ago
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