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3 years ago

Simplify x to the 5th over x to the 4th

Mathematics

1 answer:

Maru [420]3 years ago

7 0

Answer:

Step-by-step explanation:

x^5

---------

x^4

When we divide exponents with the same base, we can subtract the exponents

x^(5-4)

x^1

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

8 0

2 years ago

If 160 increased by 40 percent what is the resulting number

ANEK [815]

The resulting number would be 120

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3 years ago

Read 2 more answers

Which values are in the solution set of the compound inequality? Check all that apply. 4(x + 3) ≤ 0 or x + 1 > 3 –6 –3 0 3 8

Colt1911 [192]

we have

$4(x + 3) \leq 0$ -------> inequality 1

$x + 1 > 3$ -------> inequality 2

we know that

In this system of inequalities, for a value to be the solution of the system, it is enough that it satisfies at least one of the two inequalities.

let's check each of the values

case 1) x=-6

Substitute the value of x=-6 in the inequality 1

$4(-6 + 3) \leq 0$

$4(-3) \leq 0$

$-12 \leq 0$ -------> is ok

The value of x=-6 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

case 2) x=-3

Substitute the value of x=-3 in the inequality 1

$4(-3 + 3) \leq 0$

$4(0) \leq 0$

$0 \leq 0$ -------> is ok

The value of x=-3 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

case 3) x=0

Substitute the value of x=0 in the inequality 1

$4(0 + 3) \leq 0$