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Hunter-Best [27]
3 years ago
8

The perimeter of the rectangle is 20\text{ cm}20 cm20, start text, space, c, m, end text. One side is 6\,\text{cm}6cm6, start te

xt, c, m, end text long. What is the length of the missing side?
Mathematics
1 answer:
Tamiku [17]3 years ago
3 0

Answer: The lenght of the missing side is 4 cm

Step-by-step explanation:

The correct question is:

<em>The perimeter of the rectangle is 20cm . One side is 6cm. What is the length of the missing side?</em>

So, to answer it we have to apply the next formula:

Perimeter of a rectangle = 2 width + 2 length

Replacing with the values given: (assuming that the side given is the length of the rectangle)

20 = 2(6) + 2x

Solving for x:

20 =12 +2x

20-12 =2x

8 =2x

8/2 =x

4=x

The length of the missing side is 4 cm

Feel free to ask for more if needed or if you did not understand something.  

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Step-by-step explanation:

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Match each vector operation with its resultant vector expressed as a linear combination of the unit vectors i and j.
Cloud [144]

Answer:

3u - 2v + w = 69i + 19j.

8u - 6v = 184i + 60j.

7v - 4w = -128i + 62j.

u - 5w = -9i + 37j.

Step-by-step explanation:

Note that there are multiple ways to denote a vector. For example, vector u can be written either in bold typeface "u" or with an arrow above it \vec{u}. This explanation uses both representations.

\displaystyle \vec{u} = \langle 11, 12\rangle =\left(\begin{array}{c}11 \\12\end{array}\right).

\displaystyle \vec{v} = \langle -16, 6\rangle= \left(\begin{array}{c}-16 \\6\end{array}\right).

\displaystyle \vec{w} = \langle 4, -5\rangle=\left(\begin{array}{c}4 \\-5\end{array}\right).

There are two components in each of the three vectors. For example, in vector u, the first component is 11 and the second is 12. When multiplying a vector with a constant, multiply each component by the constant. For example,

3\;\vec{v} = 3\;\left(\begin{array}{c}11 \\12\end{array}\right) = \left(\begin{array}{c}3\times 11 \\3 \times 12\end{array}\right) = \left(\begin{array}{c}33 \\36\end{array}\right).

So is the case when the constant is negative:

-2\;\vec{v} = (-2)\; \left(\begin{array}{c}-16 \\6\end{array}\right) =\left(\begin{array}{c}(-2) \times (-16) \\(-2)\times(-6)\end{array}\right) = \left(\begin{array}{c}32 \\12\end{array}\right).

When adding two vectors, add the corresponding components (this phrase comes from Wolfram Mathworld) of each vector. In other words, add the number on the same row to each other. For example, when adding 3u to (-2)v,

3\;\vec{u} + (-2)\;\vec{v} = \left(\begin{array}{c}33 \\36\end{array}\right) + \left(\begin{array}{c}32 \\12\end{array}\right) = \left(\begin{array}{c}33 + 32 \\36+12\end{array}\right) = \left(\begin{array}{c}65\\48\end{array}\right).

Apply the two rules for the four vector operations.

<h3>1.</h3>

\displaystyle \begin{aligned}3\;\vec{u} - 2\;\vec{v} + \vec{w} &= 3\;\left(\begin{array}{c}11 \\12\end{array}\right) + (-2)\;\left(\begin{array}{c}-16 \\6\end{array}\right) + \left(\begin{array}{c}4 \\-5\end{array}\right)\\&= \left(\begin{array}{c}3\times 11 + (-2)\times (-16) + 4\\ 3\times 12 + (-2)\times 6 + (-5) \end{array}\right)\\&=\left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle\end{aligned}

Rewrite this vector as a linear combination of two unit vectors. The first component 69 will be the coefficient in front of the first unit vector, i. The second component 19 will be the coefficient in front of the second unit vector, j.

\displaystyle \left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle = 69\;\vec{i} + 19\;\vec{j}.

<h3>2.</h3>

\displaystyle \begin{aligned}8\;\vec{u} - 6\;\vec{v} &= 8\;\left(\begin{array}{c}11\\12\end{array}\right) + (-6) \;\left(\begin{array}{c}-16\\6\end{array}\right)\\&=\left(\begin{array}{c}88+96\\96 - 36\end{array}\right)\\&= \left(\begin{array}{c}184\\60\end{array}\right)= \langle 184, 60\rangle\\&=184\;\vec{i} + 60\;\vec{j} \end{aligned}.

<h3>3.</h3>

\displaystyle \begin{aligned}7\;\vec{v} - 4\;\vec{w} &= 7\;\left(\begin{array}{c}-16\\6\end{array}\right) + (-4) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}-112 - 16\\42+20\end{array}\right)\\&= \left(\begin{array}{c}-128\\62\end{array}\right)= \langle -128, 62\rangle\\&=-128\;\vec{i} + 62\;\vec{j} \end{aligned}.

<h3>4.</h3>

\displaystyle \begin{aligned}\;\vec{u} - 5\;\vec{w} &= \left(\begin{array}{c}11\\12\end{array}\right) + (-5) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}11-20\\12+25\end{array}\right)\\&= \left(\begin{array}{c}-9\\37\end{array}\right)= \langle -9, 37\rangle\\&=-9\;\vec{i} + 37\;\vec{j} \end{aligned}.

7 0
3 years ago
F(x) = –x^2+ 6<br> Find f(-3)
Lynna [10]
So if f(-3), then x = -3.


Substitute.



y = -(-3)^2 + 6

= -(9) + 6

= -3


Hope it helped
6 0
3 years ago
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