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KonstantinChe [14]
3 years ago
5

The vertices of triangle R'S'T' are R'(1, 3), S'(6, 4), and T'(4, 2). If triangle RST was dilated about the origin with a scale

factor of 3, what are the coordinates of its vertices?
Mathematics
2 answers:
Afina-wow [57]3 years ago
6 0

Answer:

R  (1/3 ,1)

S = (2,4/3)

T= (4/3,2/3)

Step-by-step explanation:

RST  becomes R'S'T'  by multiplying by 3

To get back to RST we need to divide by 3

R =(1/3, 3/3) = (1/3 ,1)

S =(6/3,4/3) = (2,4/3)

T= (4/3,2/3)

sammy [17]3 years ago
4 0

Answer:

R(1/3, 1), S(2, 4/3), T(4/3, 2/3)

Step-by-step explanation:

Dilation by a factor of 3 about the origin multiplies each coordinate by 3. To find what was multiplied by 3 to get the given coordinates, we divide the given coordinates by 3.

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the quotient is another word for division so the quotient in this equation is:

2x − 3 x ÷ 7 x2

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A school has 1,200 students. If 65% are girls, how many girls attend the school?
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Which number is irrational?
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Whats 0909090909090909090909090909+4
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Step-by-step explanation:

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2 years ago
I'm genuinely lost in any problems like this and my teachers are no help, I need help doing exponents in general, everything to
Stolb23 [73]

Answer:

See below.

Step-by-step explanation:

4)

So we have the expression:

(-6n^{-3})^2

We can use the power of a product property, where:

(ab)^n=a^n\cdot b^n

So:

=(-6)^2\cdot(n^{-3})^2

For the left, -6 squared is the same as -6 times -6. This equals positive 36.

For the right, we can use the power of a power property. The property says that:

(a^n)^k=a^{nk}

So:

(n^{-3})^2=n^{(-3)(2)}\\=n^{-6}

So, all together, we have:

=(-6)^2\cdot(n^{-3})^2\\=36n^{-6}

6)

We have the expression:

-\frac{3x^0}{x^4}

First, note that anything to the zeroth power (except for 0) is 1, thus, x^0 is also 1. Simplify:

=-\frac{3}{x^4}

And that's the simplest we can do :)

Notes for 6)

We <em>can</em> put the x^4 to the numerator. Recall that when you put an exponent to opposite side, you put a negative. In other words:

x^n=\frac{1}{x^{-n}}

And vice versa:

\frac{1}{x^{-n}}=x^n

So, we can write the above as:

=-\frac{3}{x^4}\\=-\frac{3}{x^{-(-4)}}\\=-3x^{-4}

However, traditionally, we want only positive exponents, so this wouldn't be correct.

4 0
3 years ago
Read 2 more answers
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