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3 years ago

Find the solution set of the inequality 12 - 6x > 24.

Mathematics

2 answers:

earnstyle [38]3 years ago

5 0

Answer:

x<-2

Step-by-step explanation:

Hoochie [10]3 years ago

3 0

Answer:

x < -2

Step-by-step explanation:

12 - 6x > 24 Equation

-6x > 12 Subtract 12 from both sides

x < -2 Divide both sides from -2, if you divide by any negative number in an inequality you have to switch the sign

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If you are pushing on a box with a force of 20 N and there is a force of 7 N on the box due to sliding friction, what is the net

nika2105 [10]

13 N in the direction you are pushing it.

20 N------>
<---7 N
=
13 N---->

3 0

3 years ago

Sqrtx^2+6x+9 if x≥3<br><br><br><br> please help quick idk how to do this at all

katrin [286]

Answer:

x+3

Step-by-step explanation:

sqrt(x^2+6x+9)

factor

We know this is a perfect square trinomial

sqrt(( x+3)^2)

Taking the square root of a square

x+3

7 0

3 years ago

Write an equation but do not solve to show how many letters, L, it would take for these two trophies to cost the same amount.

Juli2301 [7.4K]

Answer:

12 + 0.40L = 14.75 + 0.25L

3 0

2 years ago

What is the answer to 4.9 = 0.7m

kolbaska11 [484]

4.9 = 0.7m |(*10)
4.9 * 10 = 0.7m * 10
49 = 7m |(:7)
49/7 = 7m/7
m = 7

6 0

3 years ago

Read 2 more answers

Write the equation of the circle whose endpoints of a diameter are (2,-5) and (8,-1)

Goryan [66]

if the endpoints are there, that means that segment with those endpoints is the diameter of the circle, and that also means that the midpoint of that diameter is the center of the circle.

it also means that the distance from the midpoint to either endpoint, is the radius of the circle.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{8+2}{2}~~,~~\cfrac{-1-5}{2} \right)\implies (5,-3)\impliedby center \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{midpoint}{(\stackrel{x_1}{5}~,~\stackrel{y_1}{-3})}\qquad \stackrel{endpoint}{(\stackrel{x_2}{8}~,~\stackrel{y_2}{-1})}\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[8-5]^2+[-1-(-3)]^2}\implies r=\sqrt{(8-5)^2+(-1+3)^2} \\\\\\ r=\sqrt{3^2+2^2}\implies r=\sqrt{13} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{5}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{13}}{ r} \\[2em] [x-5]^2+[y-(-3)]^2=(\sqrt{13})^2\implies \boxed{(x-5)^2+(y+3)^2=13}$

3 0

3 years ago