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3 years ago

Please help!! ASAP thanks

Mathematics

1 answer:

AfilCa [17]3 years ago

7 0

The answer to this question should be 13. If we use the Pythagorean theorem we can enter in the values and find the missing side.

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yes was walking along the road at 5 mph and was 17 miles away from his destination. no was walking at 7 mph but was 22 miles awa

yarga [219]

17 miles / 5 mph = 3.4 hours

22miles / 7 mph = 3.14 hours

No would arrive before yes

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3 years ago

Given the diagram below, what statement can you not make?

Vaselesa [24]

Answer:

b, 5=3

Step-by-step explanation:

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3 years ago

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Are the fractions 1/5 , 5/5 , and 5/1 equivalent?Explain.

Debora [2.8K]

No, these equations are not equivalent.

1/5, or one fifth, is part of a whole. Imagine you have a pie, cut into five pieces, and your friend comes over and eats four pieces, so now you have one of the five original pieces. That's what you have here.

5/5, or five fifths, is a whole. any number divided by itself is automatically one, so it is like making another pie and cutting it into five pieces, only this time no one eats any of it because it's burned or something. At the end, you have five pieces of pie

5/1 is actually just another way of writing plain old 5. To keep the pie example rolling, you have five pies, and no one eats any of these either, so they are all yours. You have 5 pies divided between one person, so at the end of the day you have 5 whole pies.

Hope that helped!

7 0

3 years ago

Read 2 more answers

Find cos0 where 0 is the angle shown. Give an exact value, not a decimal approximation.

bogdanovich [222]

Answer:

cos(∅) = 3/5

Step-by-step explanation:

cos(∅) = adjacent/hypotenuse

We don't know what the hypotenuse is so we gotta use Pythagorean theorem to find it.

a² + b² = c²

4² + 3² = c²

√(4² + 3²) = c

c = 5 , this is our hypotenus

cos(∅) = adjacent/hypotenuse

cos(∅) = 3/5

8 0

3 years ago

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A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

4 0

3 years ago