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3 years ago

It’s my last question :(

Mathematics

1 answer:

jenyasd209 [6]3 years ago

6 0

Answer:

Perimeter = 28·√2 + 24 feet

Step-by-step explanation:

The dimensions of the initial sheet of plywood are;

The length of the sheet of plywood = 25 ft.

The width of the sheet of plywood = 14 ft.

The shape cut from each corner of the sheet of plywood = A right triangle

The leg length of each of the cut out right triangles = 7 ft.

The number of leg lengths of the right triangle cut from the length side of the initial sheet of plywood = 2

The length of the parallel sides of the remaining hexagonal piece of plywood = Initial length of the plywood - 2 × The leg length of the cut out right triangle

∴ The length of the parallel sides of the remaining hexagonal piece of plywood = 26 ft. - 2 × 7 ft. = 12 ft.

The other side length of the remaining hexagonal piece of plywood = The hypotenuse side of the cut out right triangle

The hypotenuse side of the cut out right triangle = √((7 ft.)² + (7 ft.)²) = 7·√2 ft.

∴ The other side length of the remaining hexagonal piece of plywood = 7·√2

The number of side lengths in the remaining hexagonal piece of plywood = 4

The perimeter of the remaining hexagonal piece of plywood = 2 × The length of the parallel sides + 4 × The other side lengths

∴ The perimeter of the remaining hexagonal piece of plywood = 2 × 12 ft. + 4 × 7·√2 = (28·√2 + 24) ft.

The perimeter of the remaining hexagonal piece of plywood = (28·√2 + 24) feet

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Triangle XYZ is located in the first quadrant of the coordinate plane. If this triangle is

goblinko [34]

Answer:

the answer is C. they are both positive.

Step-by-step explanation:

This is because over the y axis is the first quadrant which is always positive.

4 0

3 years ago

PLEASE HELP!!! 100 POINTS!

arlik [135]

Answer:

x = 0; y = -6; z = 1

Step-by-step explanation:

You want to find the solution to the system of equations ...

9x+y-3z=-9
10x-y+2z=8
-10x-y+4z=10

<h3>Solution</h3>

This set of equations is conveniently solved using the matrix row-reduction features of a scientific or graphing calculator, spreadsheet, or any of a number of apps or on-line calculators.

The attachment shows the solution to be ...

(x, y, z) = (0, -6, 1)

<em>Additional comment</em>

Solving a system of three or more equations "by hand" often can be done by an ad hoc process. It can be done in systematic fashion using Gauss-Jordan reduction techniques, but that often gets messy. Similarly, Cramer's Rule can be used, but that math tends to involve more arithmetic operations than are really necessary.

Adding the first equation to the other two eliminates the y-variable and reduces the system to two equations in two unknowns.

(9x +y -3z) +(10x -y +2z) = (-9) +(8) . . . . adding [1] and [2]

19x -z = -1 . . . . simplified

(9x +y -3z) +(-10x -y +4z) = (-9) +(10) . . . . adding [1] and [3]

-x +z = 1 . . . . simplified

At this point, we could graph the two equations, or we can proceed algebraically.

Adding these two equations gives ...

(19x -z) +(-x +z) = (-1) +(1)

18x = 0 ⇒ x = 0

Using the second of the reduced equations, we find ...

-x +z = 1

0 +z = 1 ⇒ z = 1

And using the second of the original equations, gives us ...

10(0) -y +2(1) = 8

-y = 6 . . . . . subtract 2

y = -6 . . . . multiply by -1

Then the solution is (x, y, z) = (0, -6, 1), as above.

6 0

1 year ago

Ples help me find slant assemtotes

FrozenT [24]

A polynomial asymptote is a function $p(x)$ such that

$\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0$

$(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}$

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form $p(x)=ax+b$ .

Ignore the negative root (we don't need it). If $y=2x-1+2\sqrt{x^2-x}$ , then we want to find constants $a,b$ such that

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

We have

$\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}$

since $x\to\infty$ forces us to have $x>0$ . And as $x\to\infty$ , the $\dfrac1x$ term is "negligible", so really $\sqrt{x^2-x}\approx x$ . We can then treat the limand like

$2x-1+2x-ax-b=(4-a)x-(b+1)$

which tells us that we would choose $a=4$ . You might be tempted to think $b=-1$ , but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of $b$ , we have to solve for it in the following limit.

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0$

$\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2$

We write

$(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}$

Now as $x\to\infty$ , we see this expression approaching $-\dfrac12$ , so that

$-\dfrac12=\dfrac{b+1}2\implies b=-2$

So one asymptote of the hyperbola is the line $y=4x-2$ .

The other asymptote is obtained similarly by examining the limit as $x\to-\infty$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

$\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0$

Reduce the "negligible" term to get

$\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0$

Now we take $a=0$ , and again we're careful to not pick $b=-1$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0$

$\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2$

$(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
 x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}$

This time the limit is $\dfrac12$ , so

$\dfrac12=\dfrac{b+1}2\implies b=0$

which means the other asymptote is the line $y=0$ .

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