Which graph is the correct shaded in the graph for the inequality provided?

Mathematics

1 answer:

KatRina [158]3 years ago

7 0

It is helpful to know several forms of the equation of a line. One that is often overlooked is the intercept form.
.. x/(x-intercept) +y/(y-intercept) = 1

The boundary conditions for your inequalities can be written as the lines
.. x/20 +y/20 = 1
.. x/24 +y/15 = 1

The first inequality will be bounded by the line that has x=20 and y=20 as its x- and y-intercepts. The second inequality will be bounded by the line with x=24 and y=15 as its x- and y-intercepts. Since the inequality conditions include the "or equal to" case, the graphed boundary line will be solid, not dashed. (All but the first graph have these lines properly shown.)

The region shaded for each inequality will be the half-plane (or its portion in the first quadrant) where the x- and y-values make the inequality true. For this problem, that is values of x and y to the left/below the line in both cases. Graph (c) shows where these "feasible regions" overlap, so is the correct choice.

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How do you write 70% as a fraction in simplest form ?

alexdok [17]

70% equals to 70/100 in fraction form.
But, you need to write it in simplest form. so, divide both of them from 10 'cause they both can divide completely by it.
70/100 = 7/10
so, your answer is 7/10

3 0

3 years ago

Don't get this

kap26 [50]

I'll talk you through it so you can see why it's true, and then
you can set up the 2-column proof on your own:

Look at the two pointy triangles, hanging down like moth-wings
on each side of 'OC'.

-- Their long sides are equal, OA = OB, because both of those lines
are radii of the big circle.

-- Their short sides are equal, OC = OC, because they're both the same line.

-- The angle between their long side and short side ... the two angles up at 'O',
are equal, because OC is the bisector of the whole angle there.

-- So now you have what I think you call 'SAS' ... two sides and the included angle of one triangle equal to two sides and the included angle of another triangle.
(When I was in high school geometry, this was not called 'SAS' ... the alphabet
did not extend as far as 'S' yet, and we had to call this congruence theorem
"broken arrow".)

These triangles are not congruent the way they are now, because one is
the mirror image of the other one. But if you folded the paper along 'OC',
or if you cut one triangle out and turn it over, it would exactly lie on top of
the other one, and they would be congruent.

So their angles at 'A' and at 'B' are also equal ... those are the angles that
you need to prove equal.

5 0

3 years ago

Divide using a common factor of 6 to find an equivalent fraction for

Savatey [412]

5: six twelfths is one half

6: eight tenths is four fifths

7: four eighths is one half

8: ten tenths is two halves

9: two sixths is one third

10: one quarter, because you divide the top and bottom number by 3

method:

take the first one - four eighths, which is 4/8