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V125BC [204]
1 year ago
11

the entire school went 250 students went to the soap box derby the math club went in 2 busses each holding 6 students

Mathematics
1 answer:
nikitadnepr [17]1 year ago
3 0

The number of students from the math club that went to the soap box derby.

<h3>How to solve algebra word problems?</h3>

From the question, we see that there are two vans and 6 students in each van.

Thus, to get the number of students from the math club that went to the soap box derby, we will solve as follows;

Number of students that went to derby = 2 × 6 = 12

Complete question is;

The entire school 250 students went to the soap box derby. The math club went in 2 vans and each van held 6 students how many students from the math club went to the soap box derby

Read more about Algebra Word Problems at; brainly.com/question/13818690

#SPJ1

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A rectangular storage container with a lid is to have a volume of 2 m3. The length of its base is twice the width. Material for
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Dimensions are 2 m by 1 meter by 1 meter,

Minimum cost is $ 18.

Step-by-step explanation:

Let w be the width ( in meters ) of the container,

Since, the length is twice of the width,

So, length of the container = 2w,

Now, if h be the height of the container,

Volume = length × width × height

2 = 2w × w × h

1 = w² × h

\implies h=\frac{1}{w^2}

Since, the area of the base = l × w = 2w × w = 2w²,

Area of the lid = l × w = 2w²,

While the area of the sides = 2hw + 2hl

= 2h( w + l)

= 2\times \frac{1}{w^2}(w+2w)

=\frac{6w}{w^2}

=\frac{6}{w}  

Since, Material for the base costs $1 per m². Material for the sides and lid costs $2 per m²,

So, the total cost,

C(w) = 1\times 2w^2+2\times 2w^2 + 2\times \frac{6}{w}

=2w^2+4w^2+\frac{12}{w}

=6w^2+\frac{12}{w}

Differentiating with respect to w,

C'(w) = 12w -\frac{12}{w^2}

Again differentiating with respect to w,

C''(w) = 12 + \frac{24}{w^3}

For maxima or minima,

C'(w) = 0

\implies 12w -\frac{12}{w^2}=0

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Hence, for width 1 m the cost is minimum,

Therefore, the minimum cost is C(1) = 6(1)²+12 = $ 18,

And, the dimension for which the cost is minimum is,

2 m by 1 meter by 1 meter.

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3 years ago
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