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3 years ago

How do you know where a line of reflection is in an art piece?

2 answers:

5 0

A line of reflection is a line that lies in a position between two identical mirror images so that any point on one image is the same distance from the line as the same point on the other flipped image. ... Divide the distance measurement by 2 to find the halfway point and mark this point with a small dot. Another name for reflection symmetry.

pogonyaev3 years ago

4 0

Answer:

Step-by-step explanation:

at the center of it's gravity

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A glacier is a giant mass of rocks and ice that flows downhill like a river. a certain glacier moves at a rate of 22 inches a da

larisa [96]

<span>To find the total distance the glacier moves we must first set up the expression for rate: 22 inches/day. Next, we convert the inches into feet and we know that 12 in = 1 ft : 22 inches/day * 1 ft/12 in = 1.83 ft/day. Finally, we multiply the number of days in year (365) by the 1.83 ft/day to calculate the total distance moved: 365 days * 1.83 ft/day = 668 ft. The glacier moved 668 feet in one year.</span>

7 0

3 years ago

Plisss help write the linear equation that vives the rule for this table

sineoko [7]

well, we know is a line, and to get the equation of any straight line all we need is two points, hmmm let's use from the table hmm say (2 , -50) and hmmm (4 , -88)

$(\stackrel{x_1}{2}~,~\stackrel{y_1}{-50})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-88}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-88}-\stackrel{y1}{(-50)}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{2}}}\implies \cfrac{-88+50}{2}\implies \cfrac{-38}{2}\implies -19$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-50)}=\stackrel{m}{-19}(x-\stackrel{x_1}{2}) \\\\\\ y+50=-19x+38\implies y=-19x-12$

5 0

2 years ago

Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x) = x2/(x4 +

ella [17]

Answer:

Given the function: $f(x) =\frac{x^2}{x^4+16}$

A geometric series is of the form of :

$\sum_{n=0}^{\infty} ar^n$

Now, rewrite the given function in the form of $\frac{a}{1-r}$ so that we can express the representation as a geometric series.

$\frac{x^2}{x^4+16}$

Now, divide numerator and denominator by $x^4$ we get;

$\frac{\frac{1}{x^2}}{1+\frac{16}{x^4}}$ = $\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2}$

Therefore, we now depend on the geometric series which is;

$\frac{1}{1+x} =\sum_{n=0}^{\infty} (-1)^n x^n$

let $x \rightarrow x^2$ then,

$\frac{1}{1+x^2} =\sum_{n=0}^{\infty} (-1)^n x^{2n}$

to get the power series let $x \rightarrow \frac{4}{x^2}$

so,

$\frac{1}{1+(\frac{4}{x^2})^2} =\sum_{n=0}^{\infty} (-1)^n (\frac{4}{x^2})^{2n}$

Multiply both side by $\frac{1}{x^2}$ we get;

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =\frac{1}{x^2} \cdot \sum_{n=0}^{\infty} (-1)^n (\frac{4}{x^2})^{2n}$

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =x^{-2} \cdot \sum_{n=0}^{\infty} (-1)^n (16)^n (x^{-2})^{2n}$

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =\sum_{n=0}^{\infty} (-1)^n (16)^n x^{-4n} \cdot x^{-2}$

Using $x^n \cdot x^m = x^{n+m}$

we have,

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =\sum_{n=0}^{\infty} (-1)^n (16)^n x^{-4n-2}$

therefore, the power series representation centered at x =0 for the given function is: $\sum_{n=0}^{\infty} (-1)^n (16)^n x^{-4n-2}$

6 0

3 years ago

Question 3 (1 point) If the following is a rhombus, find the missing measures. Find PQ. PQ=

Svetllana [295]

Answer:

Step-by-step explanation:

PQ*PQ=PQ²=(5x+16)²=25x²+160x+256

8 0

3 years ago

A package of 88 batteries costs $ 4 . 724.72 .

Vikki [24]

SOLUTION:

8 batteries = $4.72

8 ÷ 8 = 4.72 ÷ 8

1 battery = $0.59

Therefore, the cost of per battery is $0.59.
(sorry if this isn’t the correct answer)

5 0

3 years ago

Read 2 more answers