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3 years ago

The cost for 5 adults and 3 children to attend a show is $210. ( /5)

Mathematics

1 answer:

liraira [26]3 years ago

4 0

Answer: i. Cost of adult ticket = $ 30

ii. Cost of child ticket = $20

Step-by-step explanation:

Define variables:

Let x= cost of 1 adult tickets , y= cost of 1 child tickets.

Form linear equations:

5x+3y = 210 (i)

2x+5y= 160 (ii)

Multiply (i) by 2 and (ii) by 5

10x +6y = 420 (iii)

10x+25y = 800 (iv)

Eliminate (iii) from (iv)

19y = 380

⇒ y= 20

Put this in (i)

$5x+3(20)=210\Rightarrow\ 5x+60 = 210\\\\\Rightarrow\ 5x=150\\\\\Rightarrow\ x=30$

Hence, Cost of adult ticket = $ 30 and Cost of child ticket = $20

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nlexa [21]

If both printers work together. Then the number of the pamphlets will be 20x in four hours.

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Algebra is the study of mathematical symbols, and the rule is the manipulation of those symbols.

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Then the number of the pamphlets are printed by the new printer in one hour will be 4x.

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1 year ago

Write the quotient as a mixed number.<br> 63 divided by 5 = 12 R3

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3 years ago

The arithmetic sequence a; is defined by the formula:

Morgarella [4.7K]

Answer:

The sum of the first 650 terms of the given arithmetic sequence is 2,322,775

Step-by-step explanation:

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while the nth term would be ai = a(i-1) + 11

Kindly note that i and 1 are subscript of a

Mathematically, the sum of n terms of an arithmetic sequence can be calculated using the formula

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3 years ago

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3 years ago

Read 2 more answers

Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.

irina1246 [14]

Lagrange multipliers:

$L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)$

$L_x=y^2z^2+\lambda=0$
$L_y=2xyz^2+\lambda=0$
$L_z=2xy^2z+\lambda=0$
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$\lambda=-y^2z^2=-2xyz^2=-2xy^2z$

$-y^2z^2=-2xyz^2\implies y=2x$ (if $y,z\neq0$ )

$-y^2z^2=-2xy^2z\implies z=2x$ (if $y,z\neq0$ )

$-2xyz^2=-2xy^2z\implies z=y$ (if $x,y,z\neq0$ )

In the first octant, we assume $x,y,z>0$ , so we can ignore the caveats above. Now,

$x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2$

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of $f(1,2,2)=16$ .

We also need to check the boundary of the region, i.e. the intersection of $x+y+z=5$ with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force $f(x,y,z)=0$ , so the point we found is the only extremum.

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3 years ago