For the pet the fraction is 2/3 because there is a fish,cat and a dog only cats and dogs have 4 legs and the fraction for the toy or the book is 1/2 because there are 2 objects and she can only get 1

7 0

3 years ago

HELP! DUE IN A FEW HOURS

Alenkasestr [34]

Answer:

Options (C) and (E).

Step-by-step explanation:

Rotation is a rigid transformation in which size and shape of the figure and image remains the same.

That means angles and sides of the triangles given in the figure will remain same after the transformations given.

Since, the given triangle has been rotated about a point Z, each rotation will share the vertex Z.

And every figure is the transformed form of the given triangle, all four triangles will be congruent.

Options (C) and (E) are the correct options.

4 0

3 years ago

Point Q'Q ′ Q, prime is the image of Q(0,6)Q(0,6)Q, left parenthesis, 0, comma, 6, right parenthesis under the translation (x,y)

Marina86 [1]

Q is located at (0,6)

The translation rule is $(x,y) \to (x+7,y-5)$ which says to add 7 to the x coordinate and subtract 5 from the y coordinate. Doing that to (0,6) moves it to (7,1) which is where point Q' is located.

In other words, if you shift point Q(0,6) seven units to the right and five units down, then it arrives at Q ' (7,1)

<h3>Answer: (7, 1)</h3>

8 0

3 years ago

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.

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

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.

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

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

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

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

7 0

3 years ago