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3 years ago

2. Explain how the number of edges for the rectangular

Mathematics

1 answer:

Tems11 [23]3 years ago

6 0

There’s more corners and sides to a rectangular prism than a cube

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When writing a repeating decimal as a fraction does the number of repeating digits you use matter

umka21 [38]

Yes, it matters, if it is recurring or simply just irrational <span />

4 0

4 years ago

Consider the function V=g(x), where g(x) =x(6-2x)(8-2x), with x being the length of a cutout in cm and V being the volume of an

Andrej [43]

Answer:

The maximum volume of the open box is 24.26 cm³

Step-by-step explanation:

The volume of the box is given as $V=g(x)$ , where $g(x)=x(6-2x)(8-2x)$ and $0\le x\le3$ .

Expand the function to obtain:

$g(x)=4x^3-28x^2+48x$

Differentiate wrt x to obtain:

$g'(x)=12x^2-56x+48$

To find the point where the maximum value occurs, we solve

$g'(x)=0$

$\implies 12x^2-56x+48=0$

$\implies x=1.13,x=3.54$

Discard x=3.54 because it is not within the given domain.

Apply the second derivative test to confirm the maximum critical point.

$g''(x)=24x-56$ , $g''(1.13)=24(1.13)-56=-28.88\:$

This means the maximum volume occurs at $x=1.13$ .

Substitute $x=1.13$ into $g(x)=x(6-2x)(8-2x)$ to get the maximum volume.

$g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26$

The maximum volume of the open box is 24.26 cm³

See attachment for graph.

6 0

3 years ago

What is <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B5%7D%20" id="TexFormula1" title=" \frac{3}{5} " alt=" \frac{3}

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

3/5 * 15

Rewriting as

3 * 15/5

3 * 3

3 0

3 years ago

Read 2 more answers

How do I find the slope of this altitude?<br><br>I am referring to question 19.

Crazy boy [7]

The altitude's slope must be perpendicular to that of the line.

The slope of segment BC is 2/5. The slope of the altitude must be perpendicular to 2/5. Since the negative reciprocal of 2/5 is -5/2, then the slope of the altitude is -5/2.

6 0

3 years ago

Read 2 more answers

Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ

Allisa [31]

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$ . Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$ . So:

$P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002$

$P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013$

$P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082$

$P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399$

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

8 0

3 years ago