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2 years ago

Solve the following equation. Then place the correct number in the box provided. 2x - 1 ≤ x + 5

Mathematics

1 answer:

Mazyrski [523]2 years ago

7 0

$\huge\textbf{TheStarryNights }$ Hi There!

Solve the Inequality:

2x-1≤x+5

Move all the variables (in this case, x's) to the left, using the opposite operation:

2x-x-1≤5

Move -1 to the left:

2x-x≤5+1

Combine like terms:

x≤6 (Answer)

<h3>Hope it helps & enjoy your day!</h3>

$\huge\textbf{TheStarryNights}$

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Which equation represents a circle that contains the point (-5, 3) and has a center at (-2, 1)? Distance formula: vaa -02 (x - 1

natali 33 [55]

Given:

The center of the circle = (-2,1).

Circle passes through the point (-5,3).

To find:

The equation of the circle.

Solution:

Radius is the distance between the center of the circle and any point on the circle. So, radius of the circle is the distance between the points (-2,1) and (-5,3).

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$r=\sqrt{(-5-(-2))^2+(3-1)^2}$

$r=\sqrt{(-5+2)^2+(2)^2}$

$r=\sqrt{(-3)^2+(2)^2}$

On further simplification, we get

$r=\sqrt{9+4}$

$r=\sqrt{13}$

The standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where, (h,k) is the center of the circle and r is the radius of the circle.

Substitute h=-2, k=1 and $r=\sqrt{13}$ .

$(x-(-2))^2+(y-1)^2=(\sqrt{13})^2$

$(x+2)^2+(y-1)^2=13$

Therefore, the equation of the circle is $(x+2)^2+(y-1)^2=13$ .

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3 years ago

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sertanlavr [38]

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