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Yuki888 [10]
1 year ago
9

Two lines, A and B, are represented by the equations given below: Line A: 2x + 2y = 8 Line B: x + y = 4 Which statement is true

about the solution to the set of equations? It is (8, 4). It is (4, 8). There is no solution. There are infinitely many solutions.
Please Help!!!! Will Give 100 points to the Best answer.
Mathematics
1 answer:
brilliants [131]1 year ago
4 0

Line A: 2x + 2y = 8

Line B: x + y = 4


Multiply line B by 2

Line B: 2x +2y = 8


Now you have:

Line A: 2x + 2y = 8

Line B: 2x +2y = 8

--------------------------------subtract

0 = 0


Answer

There are infinitely many solutions.

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NemiM [27]
Check the picture below.

is not very specific above, but sounds like it's asking for an equation for the trapezoid only, mind you, there are square tiles too.

but let's do the trapezoid area then, 

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\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\

\bf A=\cfrac{h(a+b)}{2}\quad 
\begin{cases}
A=At\\
a=x\\
h=x\\
b=2x
\end{cases}\implies At=\cfrac{x(x+2x)}{2}
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At=\cfrac{x(3x)}{2}\implies At=\cfrac{3x^2}{2}\impliedby \textit{now, solving for \underline{x}}
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2At=3x^2\implies \cfrac{2At}{3}=x^2\implies \sqrt{\cfrac{2At}{3}}=x\implies \left( \frac{2At}{3} \right)^{\frac{1}{2}}=x

8 0
2 years ago
Josh travels frequently. For a particular airline, it takes 20 minutes for the first bag to arrive in baggage claim after a flig
Orlov [11]

Using the uniform distribution, it is found that:

A,B) 0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.

C) 20% of the time does Josh’s bag arrive in less than 22 minutes.

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

An uniform distribution has two bounds, a and b.  

The probability of finding a value of at lower than x is:

P(X < x) = \frac{x - a}{b - a}

The probability of finding a value between c and d is:

P(c \leq X \leq d) = \frac{d - c}{b - a}

The probability of finding a value above x is:

P(X > x) = \frac{b - x}{b - a}

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

  • In the graph, we have that the distribution is uniform between 20 and 30 minutes, thus a = 20, b = 30

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

Itens a and b:

  • Above 27 minutes, thus:

P(X > 27) = \frac{30 - 27}{30 - 20} = 0.3

0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.

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

Item c:

  • Less than 22 minutes, thus:

P(X < 2) = \frac{22 - 20}{30 - 20} = 0.2

0.2*100 = 20%

20% of the time does Josh’s bag arrive in less than 22 minutes.

A similar problem is given at brainly.com/question/15855314

6 0
1 year ago
Find the first partial derivative:<br><br> f(x, y) = y^5 -3xy
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Partial derivative is derivative over only 1 variable.
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2 years ago
What is the difference of the polynomials?(–2x3y2 + 4x2y3 – 3xy4) – (6x4y – 5x2y3 – y5)A–6x4y – 2x3y2 + 9x2y3 – 3xy4 + y5B–6x4y 
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We have to calculate the difference of the given polynomials, we follows as:

(-2x^{3}y^{2}+4x^{2}y^{3}-3xy^{4})-(6x^{4}y-5x^{2}y^{3}-y^{5})

After opening the brackets, the signs of all the terms changes as there is negative sign before the bracket.

=(-2x^{3}y^{2}+4x^{2}y^{3}-3xy^{4})-6x^{4}y+5x^{2}y^{3}+y^{5})

Combining all the like terms, we get as

=(-2x^{3}y^{2})+(4x^{2}y^{3}+5x^{2}y^{3})-3xy^{4}-6x^{4}y+y^{5}

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Read 2 more answers
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Answer:

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4. 3rd quartile (Q3)

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For (b). Histogram and stem & leaf. Although both usually help us understand the skewness of data distribution, however, histogram deals with frequency distribution (counts of number of occurrence) and plotted on the intervals and stem&leaf list the values.

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3 0
2 years ago
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