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3 years ago

How many Tables are needed to make a long table that will seat 22 people

Mathematics

1 answer:

nignag [31]3 years ago

6 0

How long are the tables

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Jack is purchasing a stock that pays an annual dividend of $3.42 per share. If he purchases 400 shares for $53.18 per share, wha

sergij07 [2.7K]

Answer:

1368 dollars.

Step-by-step explanation:

given that Jack s purchasing a stock that pays an annual dividend of $3.42 per share

No of shares Jack purchased = 400

Price per share = 53.18$

Amount invested by Jack in shares = $400*53.18 =21272$ dollars

Annual dividend = 3.42 per share

Total dividend = $400*3.42 = 1368$

Annual income from dividends = 1368 $

7 0

3 years ago

Find the exact length of the curve. 36y2 = (x2 − 4)3, 5 ≤ x ≤ 9, y ≥ 0

IrinaK [193]

We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: $36 y^{2} =( x^{2} -4)^3$
We divide by 36 and take the root of both sides to obtain: $y = \sqrt{ \frac{( x^{2} -4)^3}{36} }$

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: $y = \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}$

Let's leave that for the moment and look at the formula for arc length. The formula is $L= \int\limits^c_d {ds}$ where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: $ds= \sqrt{1+( \frac{dy}{dx})^2 } dx$

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here $x^2-4$ ). More formally, we can let $u=x^{2} -4$ and then consider the derivative of $u^{3/2}du$ . Either way, we obtain,

$\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}$

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
$( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}$

This means that in our case:
$ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx$
$ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx$
$ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx$
$ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx$
$ds= \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx$

Recall, the formula for arc length: $L= \int\limits^c_d {ds}$
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:

$L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x$ evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, $[(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}$

8 0

3 years ago

Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) ∈ R if and only if ad = bc. Ar

Veseljchak [2.6K]

Answer:

The given relation R is equivalence relation.

Step-by-step explanation:

Given that:

$((a, b), (c, d))\in R$

Where $R$ is the relation on the set of ordered pairs of positive integers.

To prove, a relation R to be equivalence relation we need to prove that the relation is reflexive, symmetric and transitive.

1. First of all, let us check reflexive property:

Reflexive property means:

$\forall a \in A \Rightarrow (a,a) \in R$

Here we need to prove:

$\forall (a, b) \in A \Rightarrow ((a,b), (a,b)) \in R$

As per the given relation:

$((a,b), (a,b) ) \Rightarrow ab =ab$ which is true.

$\therefore$ R is reflexive.

2. Now, let us check symmetric property:

Symmetric property means:

$\forall \{a,b\} \in A\ if\ (a,b) \in R \Rightarrow (b,a) \in R$

Here we need to prove:

$\forall {(a, b),(c,d)} \in A \ if\ ((a,b),(c,d)) \in R \Rightarrow ((c,d),(a,b)) \in R$

As per the given relation:

$((a,b),(c,d)) \in R$ means $ad = bc$

$((c,d),(a,b)) \in R$ means $cb = da\ or\ ad =bc$

Hence true.

$\therefore$ R is symmetric.

3. R to be transitive, we need to prove:

$if ((a,b),(c,d)),((c,d),(e,f)) \in R \Rightarrow ((a,b),(e,f)) \in R$

$((a,b),(c,d)) \in R$ means $ad = cb$ .... (1)

$((c,d), (e,f)) \in R$ means $fc = ed$ ...... (2)

To prove:

To be $((a,b), (e,f)) \in R$ we need to prove: $fa = be$

Multiply (1) with (2):

$adcf = bcde\\\Rightarrow fa = be$

So, R is transitive as well.

Hence proved that R is an equivalence relation.

8 0

3 years ago

Joe told Mickey he got an hourly raise at work and his new rate will be $10.25 per hour. Mickey wants to know what Joe’s hourly

insens350 [35]

<span>Joe got an hourly raise at work and his new rate will be $10.25 per hour.
</span>If r represents the amount of the raise, then the following equation can be written:
X=10.25-r
where X is <span>Joe’s hourly rate was before his raise.
Joe was paid (10.25-r) USD per hour before he got a raise.</span>

3 0

3 years ago

Read 2 more answers

Help pls thank you!!!!!!!!!!!! First to answer it RIght will receive Brainliest!!!!!!!!!!!!!!!!!!!

Verdich [7]

Answer:

Step-by-step explanation:

hope this helps!

have a great day :)

8 0

3 years ago