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3 years ago

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Which expression is equivalent to the expression below? StartFraction 6 c squared 3 c Over negative 4 c 2 EndFraction divided by

StartFraction 2 c 1 Over 4 c minus 2 EndFraction.

1 answer:

Fudgin [204]3 years ago

7 0

The given equation $(6c^2+3c)/(-4c+2)\div (2c+1)/(4c-2)$ is equivalent to the expression -3c.

Given that, the expression can be written as.

$(6c^2+3c)/(-4c+2)\div (2c+1)/(4c-2)$

By simplifying the above equation,

$\dfrac{6c^2+3c}{-4c+2}\div \dfrac{2c+1}{4c-2}$

By taking out the common terms from the equation,

$\dfrac{3c(2c+1)}{2(-2c+1)}\div\dfrac{2c+1}{2(2c-1)}$

By simplifying the above equation by cancel out the common factors.

$\dfrac{3c}{-2c+1} \div \dfrac{1}{2c-1}$

Now, by taking (-1) common from (-2c+1) we get,

$\dfrac{3c}{-1(2c-1)} \div \dfrac{1}{2c-1}$

By simplifying the above equation, we get the expression,

$-3c$

So the given equation $(6c^2+3c)/(-4c+2)\div (2c+1)/(4c-2)$ is equivalent to the expression -3c.

For more details, follow the link given below.

brainly.com/question/1301963.

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Does anyone know this and if it is extraneous or not?

Anna11 [10]

Addd 4 to both sides
$\sqrt{x+9}=5$
sqare both sides
x+9=25
minus 9 both sides
x=16

plug it in for x and see
$\sqrt{16+9}-4=1$
$\sqrt{25}-4=1$
$5-4=1$
1=1
true

not extraneous
an extraneous root would be x=-34

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3 years ago

Rectangle ABCD with coordinates A(1, 1), B(4, 1), C(4, 2), and D(1, 2) dilates with respect to the origin to give rectangle A'B'

Virty [35]

Scale factor is a ratio. The correct option is C.

<h3>How are scale drawings formed?</h3>

For a particular scale drawing, it is already specified that all the measurements' some constant scaled version will be taken. For example, let the scale be K feet to s inches.

Then it means

$\rm 1\: ft : \dfrac{s}{k}\: in.$

All feet measurements will then be multiplied by s/k to get the drawing's corresponding lengths.

The length of AB can be written as,

AB = √[(1-4)²+ (1-1)²]

AB = 3

Since the length A'B' is 6 units, therefore, the scale factor will be,

Scale factor = (Length after transformation)/(Length before transformation)

Scale factor = 6 /3 = 2

Hence, the correct option is C.

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2 years ago

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term $x^2$ , which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression: $x_v=\frac{-b}{2a}$

Since in our case $a=1$ and $b=5$ , we get that the x-position of the vertex is: $x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

$y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}$

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the $x_v=-\frac{5}{2}$ . Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

$f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0$

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

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3 years ago

Draw a line from each rectangular prism to its matching group of face shapes

IrinaK [193]

Can you put more information into this question

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3 years ago

Read 2 more answers

Please help and thank you!

Stolb23 [73]

The conversion is

$x = r \cos \theta$

$y = r \sin \theta$

So

$\dfrac y x = \tan \theta$

We're given the fairly unusual

$\theta = -\frac 5 2$

It's fairly unusual because usually an angle is given in degrees or radians, and typically the radians are a fraction times pi. We'll assume radians.

$\dfrac y x = \tan \theta = \tan(-5/2) \approx 0.747$

Ah, there's our .75. The approximate answer is

$y = 0.75 x$

Third choice.

The exact answer is

$y = x \tan( -\frac 5 2)$

also a line through the origin.

3 0

3 years ago