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Verdich [7]
3 years ago
14

Victoria was given the expression 5x - 4x^2-1+X-8x^2+7 and asked to

Mathematics
2 answers:
Nadya [2.5K]3 years ago
7 0

Answer:  -6(2x+1)(x-1)

Step-by-step explanation:

Given the expression in the form 5x-4x^2-1+x-8x^2+7, we know that Victoria can simplify it by adding the like terms:

-12x^2+6x+6

She can also factor out -6:

-6(2x^2-x-1)

Now, she needs  to factor the coefficient of the second term rewrite it as (1-2), then:

-6(2x^2+(1-2)x-1)

-6(2x^2-2x+x-1)

Now she needs to group the terms as following:

-6[(2x^2-2x)+(x-1)]

And factor out 2x:

-6[2x(x-1)+(x-1)]

Finally, she has to factor out (x-1). Therefore, she gets:

-6(2x+1)(x-1)

vfiekz [6]3 years ago
3 0

Answer:

-12x^2+6x+6

this is the answer for ap ex

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kap26 [50]

Answer:

The answer is B.

Step-by-step explanation:

You simply just simplify the expression. When you do this you get x to the third, 2x squared, and 4x. I hope this helps!!

7 0
3 years ago
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Equation editor does not include the grouping symbols “<“ and “>” that are necessary for writing a vector in component for
Scilla [17]

Answer:

{12,2}

Step-by-step explanation:

From the given graph it is clear that the initial point of the vector is (-5,0) and the terminal point (7,2).

If initial point of a vector is (x_1,y_1) and terminal point is (x_2,y_2), then

\overrightarrow{v}=(x_2-x_1)i+(y_2-y_1)j

Using this formula, we get

\overrightarrow{v}=(7-(-5))i+(2-0)j

\overrightarrow{v}=12i+2j

\overrightarrow{v}=

Using braces, we get

\overrightarrow{v}=\{12,2\}

Therefore, the required vector is {12,2}.

4 0
3 years ago
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Fiesta28 [93]

Answer:

i don't understand the question

7 0
3 years ago
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Galina-37 [17]

Answer:

C XD

Step-by-step explanation:

The altitude of a triangle goes from the vertex (corner) to the opposite side.  It must be perpendicular to the opposite side ( at a right angle)

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3 years ago
Can someone help me get the answer for this? i don’t understand it, any help is appreciated :))
Dima020 [189]

A circle is a geometric object that has symmetry about the vertical and horizontal lines through its center. When the circle is a unit circle (of radius 1) centered on the origin of the x-y plane, points in the first quadrant can be reflected across the x- or y- axes (or both) to give points in the other quadrants.

That is, if the terminal ray of an angle intersects the unit circle in the first quadrant, the point of intersection reflected across the y-axis will give an angle whose measure is the original angle subtracted from the measure of a half-circle. Since the measure of a half-circle is π radians, the reflection of the angle π/6 radians will be the angle π-π/6 = 5π/6 radians.

Reflecting 1st-quadrant angles across the origin into the third quadrant adds π radians to their measure. Reflecting them across the x-axis into the 4th quadrant gives an angle whose measure is 2π radians minus the measure of the original angle.

7 0
3 years ago
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