Wait is it not at all 14.5 or 14 1/2 Because with the math I did I got 14 1/2
14. Add the line times the outside number
X(X+4)= (2√3)^2
X^2+4X= 4√9
X^2+4X= 4times 3
X^2+4X=12
X^2+4X-12=0
(X+6)(X-2)=0
X=-6 X=2
Answe can't be negative so it's X=2
Answer:
y > (1/3)x - 3.
Step-by-step explanation:
The graph of a line can be written in y = mx + b form where b is the y-intercept and m is the slope.
In the image, the slope of the graph from the given points is 1/3.
In the image, the y-intercept is -3.
Therefore, the line is y = (1/3)x -3. However, we aren’t done yet! This is an inequality, not an equation!
We see the line isn’t dotted, so that means it must be > or <.
We substitute the point (0,0) into the line equation we got and find that 0 > (1/3)(0) - 3 = -3. Since (0,0) is part of the inequality, we have that y > (1/3)x - 3.
I hope this helps! :)
Answer:
He climed 3 steps before he started counting the steps.
Step-by-step explanation:
pls mark me as a branliest
W=mg
<span>Where: </span>
<span>Weight = mass * acceleration due to gravity </span>
<span>So let's say I want to work out my weight on the moon. I know I weigh about 70kg (which would be N), but I can't use that figure for the calculation on the moon. That is what I weigh on Earth, so let's look at the equation... </span>
<span>70kg = mass * 9.81m/s^2 </span>
<span>Where 9.81m/s^2 is the acceleration due to gravity on the surface on the earth. I want to get rid of that, so let's work out my mass by division; </span>
<span>70/9.81 = 7.14kg </span>
<span>I googled the acceleration of gravity on the Moon, which was = 1.6m/s^2 </span>
<span>Let's use that in the same equation W=mg </span>
<span>W = 7.14kg * 1.6m/s^2 = 11.42N
</span><span>On the Moon, you would weigh approximately one sixth of your weight on Earth, so if your bathroom scales tell you you weigh 120 pounds, there you would weigh 20 pounds.
</span>
<span>Moon`s gravitational pull is about one-sixth to the gravitational pull on earth hence weight on moon is about one-sixth of the weight on earth.</span>
