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3 years ago

Use the X method to find the solutions of 6x2 + 2x – 20 = 0. x = x =

2 answers:

5 0

The solution to the quadratic equation given by:
6x^2+2x-20=0
will be:
6x^2+2x-20=0
factoring the above we get:
6x^2+12x-10x-20=0
6x(x+2)-10(x+2)=0
(6x-10)(x+2)=0
hence:
6x=10
x=10/6=5/3
or
x+2=0
x=-2
Thus the answer is:
x=5/3
x=-2

Irina18 [472]3 years ago

5 0

Answer:

$x=-2$ , $x=\frac{5}{3}$

Step-by-step explanation:

$6x^2 + 2x - 20 = 0$

In the given equation all terms are divisible by 2.

$3x^2 + x - 10 = 0$

Product is 3 times -10 = -30 and sum is 1

-5 times 6 is -30

-5+6 is 1, factors are -5 and 6

Break the middle term using the factors

$3x^2-5x+6x - 10 = 0$

Group first two terms and last two terms

$(3x^2-5x)+(6x - 10) = 0$

Factor out GCF from each group

$x(3x-5)+2(3x - 5) = 0$

Factor out 3x-5

$(x+2)(3x-5)= 0$

Set each factor =0 and solve for x

$x+2=0 , x=-2$

$3x-5=0, 3x=5$

$x=\frac{5}{3}$

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3 years ago

Write the set of real numbers from - 10 through 10 in interval notation.

Klio2033 [76]

Answer:

1) $(-10,10)$ - Greater than -10 and less than 10.

2) $[-10, 10]$ - Greater than or equal to -10 and less than or equal to 10.

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4) $(-10, 10]$ Greater than -10 and less than or equal to 10.

Step-by-step explanation:

There are four form of describing that set in interval notation, which presented below:

1) $(-10,10)$ - Greater than -10 and less than 10.

2) $[-10, 10]$ - Greater than or equal to -10 and less than or equal to 10.

3) $[-10, 10)$ - Greater than or equal to -10 and less than 10.

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3 years ago

Composite function is the result of using one function as the input for another function.

ipn [44]

Answer:

True

Step-by-step explanation:

That is the definition of composite function

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3 years ago

Solve the following system of equations. Write each of your answers as a fraction reduced to lowest terms. In other words, write

morpeh [17]

Answer:

x = 57/28

y = -95/84

z = 97/168

Step-by-step explanation:

Use the application in the next link: https://www.zweigmedia.com/RealWorld/tutorialsf1/scriptpivotold.html

Start with the expanded array:

$\left[\begin{array}{cccc}1&5&8&1\\3&2&2&5\\-2&-7&2&5\\\end{array}\right]$

then using the tool provided, make row operations until you find the solution:

r2 = r2-3r1

$\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\-2&-7&2&5\\\end{array}\right]$

r3 = r3+2r1

$\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\0&3&18&7\\\end{array}\right]$

r2 = r2*(-1/13)

$\left[\begin{array}{cccc}1&5&8&1\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]$

r1 = r1- r2*5

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]$

r3 = r3+ r2*-3

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&168/13&97/13\\\end{array}\right]$

r3 = r3*13/168

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&1&97/168\\\end{array}\right]$

r2 = r2- r3*22/13

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]$

r2 = r2+ r3*6/13

$\left[\begin{array}{cccc}1&0&0&57/28\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]$

Here you have a reduced array an therefore the answers to each variable are on each row:

$\left[\begin{array}{c}x\\y\\z\end{array}\right]$

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3 years ago

Find the surface area of the composite solid and round to the nearest tenth. Explain your answer.

Orlov [11]

Well, notice the composite is really just 4 triangles atop sitting on top of 4 rectangles, and all of them area stacked up at the edges.

so, for the rectangle's sides,

front and back are two 6x3 rectangles

left and right are two 6x3 rectangles

the bottom part is a 6x6 rectangle

now, we don't include the 6x6 rectangle that's touching the triangles, because that's inside area, and is not SURFACE area, so we nevermind that one.

now, the triangles are just four triangles with a base of 6, and a height of 4, in red noted there.

so, just get the area of all those rectangles and the triangles, sum them up and that's the surface area of the composite,

$\bf \stackrel{lateral~rectangles}{4(6\cdot 3)}~~+~~\stackrel{bottom~rectangle}{(6\cdot 6)}~~+~~\stackrel{triangles}{4\left( \cfrac{1}{2}(6)(4) \right)}$

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3 years ago