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dezoksy [38]
2 years ago
6

A school graduation class wants to hire buses and vans for a trip to Jasper National Park. Each bus

Mathematics
1 answer:
BARSIC [14]2 years ago
6 0
Let the school hire x buses and y vans.

A bus can hold 40 students and 3 teachers.
A van can hold 8 students and 1 teacher.

The number of students riding in  buses and vans is at least 400, therefore
40x + 8y ≥ 400          (1)

The number of teachers riding in buses and vans is at most 36, therefore
3x + y ≤ 36                (2)

Write (1) and (2) as
y ≥ 50 - 5x                 (3)
y ≤ 36 - 3x                 (4)

The equality portion of the solution of (3) and (4) is
36 - 3x = 50 - 5x
2x = 14
x = 7   =>  y = 36 - 3*7 = 15

A graph of the inequalities indicates the acceptable solution in shaded color, as shown below.

The minimum cost of renting buses and vans is
7*$1200 + 15*$100 =  $9900

Answer: The minimum cost is $9,900

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For this case, the first thing we must do is define variables.

We have then:

t: number of hours

F (t): total charge

We write the function that models the problem:

F (t) = 6t + b

Where,

b: represents an initial fee.

We must find the value of b.

For this, we use the following data:

Her total fee for a 4-hour job, for instance, is $ 32.

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From here, we clear the value of b:

32 = 24 + b

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7 0
2 years ago
A fuel is purified by passing it through a clay pipe. Each foot of the clay pipe removes a fixed percentage of impurities in the
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2 years ago
which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater
Nutka1998 [239]

Answer:

\frac{\sqrt[3]{16y^4}}{x^2}

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices \sqrt[n]{a}  = a^{\frac{1}{n}}

So,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } gives

\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}

Also from laws of indices

(ab)^n = a^nb^n

So, the above expression can be further simplified to

\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}

Multiply the exponents gives

\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

Substitute 2^5 for 32

\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

From laws of indices

\frac{a^m}{a^n} = a^{m-n}

This law can be applied to the expression above;

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})} becomes

2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}

Solve exponents

2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}

From laws of indices,

a^{-n} = \frac{1}{a^n}; So,

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}} gives

\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}

The expression at the numerator can be combined to give

\frac{(2y)^{\frac{4}{3}}}{x^2}

Lastly, From laws of indices,

a^{\frac{m}{n} = \sqrt[n]{a^m}; So,

\frac{(2y)^{\frac{4}{3}}}{x^2} becomes

\frac{\sqrt[3]{(2y)}^{4}}{x^2}

\frac{\sqrt[3]{16y^4}}{x^2}

Hence,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } is equivalent to \frac{\sqrt[3]{16y^4}}{x^2}

8 0
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Every month your balance includes the original amount (100%) and the added the monthly interest  (1.42%)  so each  month  the balance will be 101.42%  of pior month's balance move  the decimal point two points two places to the left to make that into a decimal 

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8 0
2 years ago
What angle does an arc 6.6cm in length subtends at the centre of a circle of radius 14cm. Use π = 22/7)
aniked [119]

Answer:

STEP 1: Find the circumference:

Circumference = 2πr

Circumference = 2π(14) = 28π cm

............................................................................................

STEP 2: Find the length of the arc:

Length of the arc = 36/360 x 28π

Length of the arc = 8.8 cm

.............................................................................................

Answer: The length of the arc is 8.8 cm

............................................................................................

hope it helpssss

Mark it as brilliant answer plzzz

ФωФ

4 0
2 years ago
Read 2 more answers
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