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2 years ago

Andre draws a tape diagram for 3÷2/3

Mathematics

1 answer:

bija089 [108]2 years ago

4 0

Answer:

No I do not agree with Andre says that 3 divided by 2/3 Solving for 3 ÷ 2/3

3 ÷ 2/3

= 3 × 3/2

= 9/2

= 4 1/2

Therefore, 3 divided by 2/3 is 4 1/2

Andre's reasoning is wrong.

Step-by-step explanation:

It is incorrect

Let's divide and compare the sum

3 : 2/3 = 3*3/2 = 9/2 = 4 1/2

and

There are four 2/3 fractions and one 1/3 fraction

The 1/3 fraction is 1/2 of 2/3 fraction so instead of counting 1/3 as 1/3 it should be counted as 1/2

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Please help solve the problem below.

ivann1987 [24]

Answer:

1. x = 2

2. x = 5

3. x = 8

4. x = 28

5. x = 13

6. x = 11

7. x = 0

8. x = 72

9. x = 6

10. x = 38

Step-by-step explanation:

hope that helps!

1. subtract 6 on both sides

2. add 3 on both sides

3. divide both sides by 2

4. multiply both sides by 2

5. add 8 on both sides

6. subtract 9 on both sides

7. add 5 on both sides

8. multiply both sides by 6

9. divide both sides by 4

10. add 18 on both sides

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3 years ago

Are the graphs 8x+3y=6 and 8x-3y=6 perpendicular

Alla [95]

Lines are perpendicular if the slope of one is the negative reciprocal of the other. In this case the slopes are -8/3 and 8/3, so while negative they are not reciprocal. So no, the two lines intersect but are not perpendicular.

4 0

3 years ago

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:

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

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

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2 years ago

A conjecture and the flowchart proof used to prove the conjecture are shown.

dexar [7]

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2 years ago

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Serggg [28]

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