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3 years ago

If ABCD is dilated by a factor of 3, the

Mathematics

1 answer:

Vlad [161]3 years ago

7 0

Answer:

(6,-6)

Step-by-step explanation:

First let's identify the current coordinates of D

It appears that D is located at (2 , -2)

Now let's find the coordinate of D if it were dilated by a scale factor of 3.

To find the coordinates of a point after a dilation you simply multiply the x and y values of the pre image coordinates by the scale factor

In this case the scale factor is 3 and the coordinates are (2,-2)

That being said let's apply the dilation rule

Current coordinates: (2,-2)

Scale factor:3

Multiply x and y values by scale factor

(2 * 3 , -2 * 3) --------> (6 , -6)

The coordinates of D' would be (6,-6)

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rosijanka [135]

The buyer will be paying taxes for 185 days, so will pay

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3 years ago

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

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1 year ago

The triangles shown are similar. Find x and the measure of side DE.<br> Help pls. ?

aleksandr82 [10.1K]

If they are similar, their sides are proportional. Compare 3/5 ratio is the same as x-4/x

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2 years ago

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vampirchik [111]

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2 years ago

I don’t understand what this question is asking for

Elden [556K]

Answer:

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Step-by-step explanation:

The question is asking you to compare the measures of two inscribed angles. Each of the inscribed angles intercepts the circle at points A and B, which are the endpoints of a diameter.

<h3>applicable relations</h3>

Several relations are involved here.

The measures of the arcs of a circle total 360°
A diameter cuts a circle into two congruent semicircles
The measure of an inscribed angle is half the measure of the arc it intercepts

<h3>application</h3>

In the attached diagram, we have shown inscribed angle ADB in blue. The semicircular arc it intercepts is also shown in blue. A semicircle is half a circle, so its arc measure is half of 360°. Arc AEB is 180°. That means inscribed angle ADB measures half of 180°, or 90°. (It is shown as a right angle on the diagram.)

If Brenda draws angle AEB, it would look like the angle shown in red on the diagram. It intercepts semicircular arc ADB, which has a measure of 180°. So, angle AEB will be half that, or 180°/2 = 90°.

The question is asking you to recognize that ∠ADB = 90° and ∠AEB = 90° have the same measure.

m∠AEB = m∠ADB

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<em>Additional comment</em>

Every angle inscribed in a semicircle is a right angle. The center of the semicircle is the midpoint of the hypotenuse of the right triangle. This fact turns out to be useful in many ways.

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2 years ago