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2 years ago

Can someone help for my math test? its 20 points and ill be giving brainiest!

Mathematics

2 answers:

Ludmilka [50]2 years ago

8 0

Answer: 20

Part 1:

You want to make a rectangular dog pen using 20 yards of fencing. The side lengths must be in whole yards. Create as many different rectangular pens as you can.

Part 2:

Which dog pen gives your dog the most space to run around and play in? Write a sentence explaining how you know.

Part 3:

You want to build the rectangular dog pen with 20 yards of fencing against your house which is 20 yards wide. Which dimensions will give you the most space for your dog?

Part 4:

One of your friends just got a puppy. She is going to buy fencing with her parents this weekend to build her own pen. Explain to her how to design the fence so that her puppy will have the biggest pen possible, as well as how you know it will be the biggest pen possible.

weqwewe [10]2 years ago

4 0

Answer:

the length in fraction form would be 16 1/2 rounded to the nearest 10 would be 20

Step-by-step explanation:

because 16.5=16 1/2

16.5=17

17=20

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11.REINFORCE The lengths of three sides of a triangle are 2x, 3x - 4, and x+4. Find a value of x that makes the triangle equilat

mash [69]

Answer:

x=4

Step-by-step explanation:

It suffices to equate any two sides and then check whether the solution is valid for the third side.

Taking 2x and 3x-4 and equating:

2x = 3x-4

x = 4

This means the sides are both of length 8. Now check whether for x=4 the third side is also of length 8:

x+4=4+4=8

which indeed works out confirming x=4 being a valid solution

5 0

3 years ago

Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2

iren2701 [21]

Hello there, hope I can help!

I assume you mean $L\left\{\frac{ekt+e-kt}{2}\right\}$
With that, let's begin

$\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \ L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}$

$\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}$
$\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b$
$L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}$
$\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}$

$L\left\{t\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \ L\left\{t\right\}=\frac{1}{s^2}$

$L\left\{\frac{e}{2}\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \ L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \ \frac{e}{2s}$

$\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}$

$\frac{ek}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{ek\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{ek}{2s^2}$

$\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{k\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{k}{2s^2}$

$\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}$

Hope this helps!

3 0

3 years ago

A car rental company charges a one-time application fee of 40 dollars, 55 dollars per day, and 13 cents per mile for its cars. A

liberstina [14]

Answer:

A) C(d,m) = 40 + 55d + 0.13m

B) $448

Step-by-step explanation:

Let 'd' be the number of days and 'm' the number of miles driven.

A) The cost function that describes a fixed amount of $40, added to a variable amount of $55 per day (55d) and a variable amount of 13 cents per mile (0.13m) is:

$C(d,m) = 40 +55d +0.13m$