All categories

3 years ago

Solve for x. the answer must be simplified. −36x<8

Mathematics

2 answers:

Degger [83]3 years ago

8 0

Its -.222222222222 if you even go to calculator thats the answer

STatiana [176]3 years ago

3 0

-36x<8 divide both sides by -36 (reversing the direction of the inequality sign because of division by a negative number)

x>-8/36 the largest common factor of 8 and 36 is 4 so divide numerator and denominator by this factor, 8 to simplify the fraction

x>-2/9 or in proper notation, x=(-2/9, +oo)

You might be interested in

Complete the point-slope equation of the line through -6 comma 5 and -3 comma -3

Anna007 [38]

Step-by-step explanation:

<h3>Given points </h3>

A (-6,5)
B (-3,-3)

${\boxed{\sf Slope_{(m)}=\dfrac{y _2-y_1}{x_2-x_1}}}$

Substitute the values

$\qquad {:}\longrightarrow\mathbb m=\dfrac {-3-5}{-3+6}$

$\qquad {:}\longrightarrow\mathbb m=\dfrac {-8}{3}$

$\therefore\sf Slope=\dfrac {-8}{3}$

5 0

3 years ago

NO LINKS!!! Please help me with these problems

nadezda [96]

<h3>Answers:</h3>

7) Center= (-1,2) Radius= $\boldsymbol{\sqrt{8}}$ Equation: $(x+1)^2+(y-2)^2 = 8$

8) Center= (3,13) Radius= 13 Equation: $(x-3)^2+(y-13)^2 = 169$

=========================================================

Explanation:

Problem 7

Let's find the distance from (-1,2) to (-3,4)

$(x_1,y_1) = (-1,2) \text{ and } (x_2, y_2) = (-3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-1-(-3))^2 + (2-4)^2}\\\\d = \sqrt{(-1+3)^2 + (2-4)^2}\\\\d = \sqrt{(2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\$

This is the radius because it stretches from the center to a point on the circle, so $r = \sqrt{8}$

Squaring both sides will get us $r^2 = 8$

One useful template for a circle is the equation $(x-h)^2+(y-k)^2 = r^2\\\\$

(h,k) is the center

r is the radius

Let's plug in the given center (h,k) = (-1,2) and the r^2 value we found earlier.

$(x-h)^2+(y-k)^2 = r^2\\\\(x-(-1))^2+(y-2)^2 = 8\\\\(x+1)^2+(y-2)^2 = 8\\\\$

You can confirm this by using a tool like Desmos. See below.

------------------------------------------------------------------------

Problem 8

The endpoints of the diameter are (-2,1) and (8,25)

The center is the midpoint of these endpoints.

The midpoint of the x coordinates is (-2+8)/2 = 3

The midpoint of the y coordinates is (1+25)/2 = 13

The center is (h,k) = (3,13)

Now find the distance from the center to one of the points on the circle, let's say to (8,25)

$(x_1,y_1) = (3,13) \text{ and } (x_2, y_2) = (8,25)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-8)^2 + (13-25)^2}\\\\d = \sqrt{(-5)^2 + (-12)^2}\\\\d = \sqrt{25 + 144}\\\\d = \sqrt{169}\\\\d = 13\\\\$