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2 years ago

Which are correct representations of the inequality –3(2x – 5) < 5(2 – x)? Select two options.

Mathematics

1 answer:

chubhunter [2.5K]2 years ago

4 0

The correct statements are options C and option D.

- 6x + 15 < 10 - 5x ⇒ 3rd answer

An open circle is at 5 and a bold line starts at 5 and is pointing to the right.

<h3 /><h3>What is inequality?</h3>

The relation between two expressions that are not equal, employing a sign such as ≠ ‘not equal to’, > ‘greater than, or < ‘less than.

The inequality is -3(2x - 5) < 5(2 - x)

At first, simplify each side

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

Remember (-)(-) = (+)

-3(2x - 5) = - 6x + 15

5(2 - x) = 5(2) + 5(-x)

Remember (+)(-) = (-)

5(2 - x) = 10 - 5x

- 6x + 15 < 10 - 5x

Subtract 15 from both sides

- 6x < -5 - 5x

Add 5x to both sides

- x < - 5

Remember the coefficient of x is negative, then when you divide both sides by it you must reverse the sign of inequality

The coefficient of x is -1

Divide both sides by -1

x > 5

Therefore the correct statements are options C and option D.

- 6x + 15 < 10 - 5x ⇒ 3rd answer

An open circle is at 5 and a bold line starts at 5 and is pointing to the right.

To know more about inequality follow

brainly.com/question/24372553

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Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)

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The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

This is the diameter

To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

This is the standard equation of the sphere

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3 years ago