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2 years ago

Please help, this is for a final grade and it affects how I place

Mathematics

1 answer:

saw5 [17]2 years ago

4 0

No solution
y=3x-3
-6x+2y= -4
-6x+2(3x-3)= -4
-6x+6x-6= -4
Cancel -6x+6x, since it equals 0, you are left with -6= -4. No solution.

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What is the answer to this?

Dmitry [639]

Answer:

Step-by-step explanation:

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2 years ago

The figure shows the dimension of a city park in feet what is the area of the park?

salantis [7]

The answer would b C. 1,208 ft2

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2 years ago

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11Alexandr11 [23.1K]

100,435 hope this helps you

7 0

3 years ago

Find n. Perimeter = 86 in.

FromTheMoon [43]

Let the length of rectangle be L and the width of rectangle be W.
Since length exceeds the width by 25 inches, length will be
L = W + 25

Now the perimeter, P, is given by
P = 2(L + W)
Substituting L = W + 25 in the above equation,
P = 2(W + 25 + W)
P = 2(2W + 25)
P = 4W + 50
But P = 86 inches
P = 4W + 50 = 86
4W = 86 - 50 = 36
W = 36/4 = 9

Hence, width W = 9 inches.
Length L = W + 25 = 9 + 25 = 34 inches.

5 0

3 years ago

Mrs. Del Pup bought 20 yards of border for her classroom bulletin board.

KengaRu [80]

A rectangle with dimensions $x$ and $y$ has perimeter 20 if

$2(x+y)=20 \iff x+y=10$

and we can deduce

$y=10-x$

So, the dimensions of the rectangle must be $x$ and $10-x$ , where x ranges from 0 to 10 (both extremes are excluded, otherwise you'd have a degenerate rectangle, which is actually a segment).

So, all the possible areas are given by the product of the dimensions, i.e.

$x(10-x) = -x^2+10x$

The function $f(x)=-x^2+10x$ represents a parabola, facing downwards, and thus it admits a maximum, which is its vertex.

In particular, the vertex of this parabola is given by

$x=\dfrac{-b}{2a}=\dfrac{-10}{-2}=5$

And it yields an area of

$f(5)=-25+50=25$

(a) So, she can frame a maximum area of 25 squared yards.

(b)The dimensions of the rectangle with greatest area are $x=5$ and $y=10-x=10-5=5$ , so it's actually a square.

This is a well known theorem: if you fix the perimeter, the rectangle with the largest area is the square yielding that perimeter.

(c) If we increase both dimensions by 2 yards, our 5x5 square would become a 7x7 square...

(d) ...and the new area would be $7^2=49$ squared yards. So, the new area is $49-25=26$ squared yeards more than the old one.

6 0

2 years ago