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Kitty [74]
3 years ago
9

Let u=(5,-12) and c=-3 what is || cu || ?A.-39 B.21 C.39 D.51 i need help badly

Mathematics
2 answers:
OLEGan [10]3 years ago
6 0
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3184618

——————————

\mathsf{c}\cdot \overrightarrow{\mathsf{u}}=\mathsf{-\,3\cdot (5,\,-12)}


Therefore,

\left\|\mathsf{c}\cdot \overrightarrow{\mathsf{u}}\right\|=\mathsf{\left\|-\,3\cdot (5,\,-12)\right\|}\\\\
\left\|\mathsf{c}\cdot \overrightarrow{\mathsf{u}}\right\|=\mathsf{\left|-\,3\right|\cdot \left\|(5,\,-12)\right\|}\\\\
\left\|\mathsf{c}\cdot \overrightarrow{\mathsf{u}}\right\|=\mathsf{3\cdot \sqrt{5^2+(-12)^2}}

\left\|\mathsf{c}\cdot \overrightarrow{\mathsf{u}}\right\|=\mathsf{3\cdot \sqrt{25+144}}\\\\
\left\|\mathsf{c}\cdot \overrightarrow{\mathsf{u}}\right\|=\mathsf{3\cdot \sqrt{169}}\\\\
\left\|\mathsf{c}\cdot \overrightarrow{\mathsf{u}}\right\|=\mathsf{3\cdot \sqrt{13^2}}\\\\
\left\|\mathsf{c}\cdot \overrightarrow{\mathsf{u}}\right\|=\mathsf{3\cdot 13}

\left\|\mathsf{c}\cdot \overrightarrow{\mathsf{u}}\right\|=\mathsf{39}\quad\longleftarrow\quad\textsf{this is the answer (option C. 39)}


I hope this helps. =)

Degger [83]3 years ago
4 0

Answer:39 is the awnser

Step-by-step explanation:

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A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six
Zarrin [17]

Answer:

The volume of the box is maximized when x = 0.53 cm

Therefore,  x = 0.53 cm  and the Maximum volume = 1.75 cm³

Step-by-step explanation:

Please refer to the attached diagram:

The volume of the box is given by

V = Length \times Width \times Height \\\\

Let x denotes the length of the sides of the square as shown in the diagram.

The width of shaded region is given by

Width = 3 - 2x \\\\

The length of shaded region is given by

Length = \frac{1}{2} (5 - 3x) \\\\

So, the volume of the box becomes,

V =  \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V =  \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V =  \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V =  \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\

Take the derivative of volume and set it to zero.

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We may solve the quadratic equation using the quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The values of coefficients a, b, c are

a = 18 \\\\b = -38 \\\\c = 15 \\\\

Substituting the values into quadratic formula yields,

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Volume of box when x = 1.59:

V =  \frac{1}{2} (5 - 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\

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V =  \frac{1}{2} (5 - 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3

As you can see, the volume of the box is maximized when x = 0.53 cm

Therefore,  x = 0.53 cm  and the Maximum volume = 1.75 cm³

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Step-by-step explanation:

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