Answer:
2000 L
Step-by-step explanation:
There are 1250 L of water in a tank at present. If the tank is 0.625 full, what is the capacity of the tank?
The simple solution is:
1250 L ÷ 0.625 = 2000 L
The algebraic solution is:
Let <em>c</em> equal the capacity of the tank.
Therefore, <em>c</em> × 0.625 = 1250.
Divide both sides by 0.625:
<em>c</em> × 0.625 ÷ 0.625 = 1250 ÷ 0.625
And simplify:
<em>c</em> = 1250 ÷ 0.625
<em>c</em> = 2000
No answer is possible until you tell us something about the sphere. Anything. Is it bigger than a breadbox ? Smaller than a teardrop ? We can get really mathematical about it if you'd share the sphere's radius, or diameter, or surface area. But we do need SOMEthing.
Answer:
Step-by-step explanation:
lm^2+n^2|
(-5)^2+(3)^2= 227
Answer:x=90
Step-by-step explanation:100+2x=10+3x
100-10=3x-2x
90= x
Answer:
see below
Step-by-step explanation:
I find it convenient to use a graphing calculator to show the graph of many expressions, equations, and inequalities. See below for a graph.
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If you want to graph this by hand, it is convenient to start by graphing the "equal to" case, the line ...
-x + y = 5
or
y = x + 5
This is a line with a slope of +1 and a y-intercept of (0, 5). Since the "or equal to" case is included, the graph of the line is a solid line.
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You can then locate one of the variables, preferable one with a positive coefficient. (y has a coefficient of +1) Notice its relation to the comparison symbol. Here, we have ( ) +y ≤ ( ). That is, values of <em>y less than</em> {something} will be part of the solution set. This tells you the shading is <em>below the line</em>. (The shading indicates values of x and y that are part of the solution.)
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<em>Additional comments</em>
Had the relation been y > {something}, shading would be <em>above the line</em>. You will note that the inequality can also be written ...
y ≤ x + 5
We can also look at the x-variable to determine where the shading should be. This inequality has the form {something} ≤ x, so x-values <em>greater</em> than those on the boundary line are part of the solution. Since the slope of the line is positive, "right of the line" is the same as "below the line" as far as shading is concerned.
Of course, if the relation had been < instead of ≤, the boundary line would be <em>dashed</em> (not solid) and considered to be <em>not part of the solution</em>.