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3 years ago

A restaurant manager states the number of customers that enter the

Mathematics

2 answers:

Arte-miy333 [17]3 years ago

7 0

Answer:

Yes, that is true, and x=45.

Step-by-step explanation:

Yes, that's how you set up the equation I also listed the sets to solve it below:

3x+15=2x+60

-2x. -2x

--------------------

x+15=60

-15 -15

------------------

x=45

You can also plug it in to check your work.

3(45)+15=2(45)+60

135+15=90+60

150=150

konstantin123 [22]3 years ago

6 0

Yes it is true because if 3x +15=2x+60 it is true

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What is the perimeter of the figure

Brums [2.3K]

Answer:

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides. x is in this case the length of the rectangle while y is the width of the rectangle. The perimeter, P, is: P=x+x+y+y.

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2 years ago

Could anyone help me out, it’s would mean a lot.

SIZIF [17.4K]

Answer:

$x = 19$

Step-by-step explanation:

If two lines are perpendicular, they create 4 90° angles. Meaning, in this case, that m<DBC = 90 and m<DBA = 90.

We're given that m<DBE = 2x - 1 and that m<CBE = 5x - 42.

The sum of angles DBE and CBE = m<DBC = 90°

We can add the two angles and set it equal to 90 to find x

$2x - 1 + 5x - 42 = 90\\7x - 43 = 90\\7x = 133\\x = \frac{133}{7} =19$

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3 years ago

43 divided by 982 with remainder

Marina CMI [18]

The answer to this question is 22.83 but if you want it rounded its 22.84

4 0

2 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

Does this graph represent a function? Why or why not?

malfutka [58]

Answer:

Yes.

Step-by-step explanation:

It is a function because it passes the vertical line test. Each x-value has just one y-value. If the relation had any x-values that had several y-values, it would just be a relation. But this one is a function.

Hope this helps!

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2 years ago