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matrenka [14]
3 years ago
5

Which combination of transformations can you use to show that one shape is congruent to another? dilations and reflections/ refl

ections and stretches /rotations and translations/ reflections, dilations, and translations /rotations, translations, and dilations
Mathematics
1 answer:
Ganezh [65]3 years ago
8 0
The correct answer to this is:
ROTATIONS AND TRANSLATIONS

Reflection and rotation can also be used but since their both grouped with dilations, they can't be used. When dilation is used, the size of the shape is changed and congruence won't be possible anymore . That's why the answer is the first option.
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Find f(2) for the following functions: f(x) = x - 14
BaLLatris [955]

Answer:

<h2>f(2) = - 12</h2>

Step-by-step explanation:

f(x) = x - 14

To find f(2) substitute the value of x that's 2 into f(x). That is for every x in f(x) replace it with 2

We have

f(2) = 2 - 14

We have the final answer as

<h3>f(2) = - 12</h3>

Hope this helps you

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3 years ago
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just olya [345]
X = 19.

Explanation:

This expression is vertical from the angle 135, meaning it is congruent. With that being said, we would need to find what 7 multiplies by to give us 2 less than 135, which would be 133. Because 133 divided by 7 equals 19, x = 19. If we plug this in, we get the following equation: 7(19) + 2 = 135.
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2 years ago
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Simplify. Your answer should contain only positive exponents with no fractional exponents in the denominator.
mart [117]

Answer:

\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}.

Step-by-step explanation:

The given expression is  

\dfrac{3y^{\frac{1}{4}}}{4x^{-\frac{2}{3}}y^{\frac{3}{2}}\cdot 3y^{\frac{1}{2}}}

We need to simplify the expression such that answer should contain only positive exponents with no fractional exponents in the denominator.

Using properties of exponents, we get

\dfrac{3}{4\cdot 3}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}}    [\because a^ma^n=a^{m+n}]

\dfrac{1}{4}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{2}}

\dfrac{1}{4}\cdot \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{y^{2}}         [\because a^{-n}=\dfrac{1}{a^n}]

\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}

We can not simplify further because on further simplification we get negative exponents in numerator or fractional exponents in the denominator.

Therefore, the required expression is \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}.

5 0
3 years ago
6% of $35 is $_____?
Molodets [167]

answer: 2.1%

p/w=%/100

?/35=6/100

35×6=210

210÷100

2.1

Hope this helps! Plz mark as brainliest! :)

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3 years ago
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