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DaniilM [7]
2 years ago
10

First I will build the envelope. For this, I am using a rectangular piece of material 28m long and 12m wide. Out of this materia

l, I have to cut patches that are 1/12 the length and 2/15 the width of the whole material. What is the area of a patch and how many such patches can I make out of the whole material?
Mathematics
1 answer:
RUDIKE [14]2 years ago
3 0

Answer:

3.728 cm²

90

Step-by-step explanation:

To built the envelope I am using a rectangular piece of material 28 m long and 12 m wide.  

Out of this material, I have to cut patches that are 1/12 the length and 2/15 the width of the whole material.

So, the length of the patches is 28 \times \frac{1}{12} = 2.33 m and the width of the patches is 12 \times \frac{2}{15} = 1.6 m.

Therefore, the area of the patches is (1.6 × 2.33) = 3.728 m² each. (Answer)

Now, the area of the rectangular piece of material is (28 × 12) = 336 m²

Hence, \frac{336}{3.728} = 90.37 ≈ 90 such patches can be made out of the whole material. (Answer)

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Answer:

There were 20 picks total

Step-by-step explanation:

We can use a proportion to solve this problem.  Put the orange picks on top and the total picks on bottom

3 orange         12 orange

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2 years ago
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How to solve this problem
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Lubov Fominskaja [6]

Answer:

The graph of the function f(x)=\frac{1}{2}x^{2}-4x+5 has a minimum located at (4,-3)

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

f(x)=a(x-h)^{2}+k

where

a is a coefficient

(h,k) is the vertex of the parabola

If a > 0 the parabola open upward and the vertex is a minimum

If a < 0 the parabola open downward and the vertex is a maximum

In this problem

The coefficient a must be positive, because we need to find a minimum

therefore

Check the option C and the option D

Option C

we have

f(x)=\frac{1}{2}x^{2}-4x+5

Convert to vertex form

f(x)-5=\frac{1}{2}x^{2}-4x

Factor the leading coefficient

f(x)-5=\frac{1}{2}(x^{2}-8x)

f(x)-5+8=\frac{1}{2}(x^{2}-8x+16)

f(x)+3=\frac{1}{2}(x^{2}-8x+16)

f(x)+3=\frac{1}{2}(x-4)^{2}

f(x)=\frac{1}{2}(x-4)^{2}-3

The vertex is the point (4,-3) ( is a minimum)

therefore

The graph of the function f(x)=\frac{1}{2}x^{2}-4x+5 has a minimum located at (4,-3)

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