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Oduvanchick [21]
3 years ago
9

What transformations (listed in order) were used to move the figure on the left to the one on the right? A) translate left 8 and

up 2 B) translate left 1, translate up 2, and reflect over y-axis C) translate left 1, translate up 2, and reflect over x-axis D) reflection over y-axis, translate left 1 and translate down 2
Mathematics
2 answers:
Paul [167]3 years ago
7 0
The correct order is reflection over y-axis, translate left 1 and translate down 2
KIM [24]3 years ago
6 0

i found that D was the correct answer. have a good day

          answer: D

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alexandr1967 [171]

Answer:

I KNOW FOR SURE

Step-by-step explanation:

if that's all the question is then I definitely know the answer ;)

if you forgot to write the rest then hmu haha

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In a game of luck, a turn consists of a player rolling 7 six-sided dice and counting how many of the dice
harkovskaia [24]

Answer: Binomial

Step-by-step explanation:

1 / 3

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2 years ago
What is 2/7 + 6/10 in fraction form
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31/35

Step-by-step explanation:

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Which ratio can<br> form a proportion<br> with 2/3?
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3 years ago
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when 6 is subtracted from the square of a number, the result is 5 times the number. Find the negative solution.
sveta [45]

When 6 is subtracted from the square of a number, the result is 5 times the number, then the negative solution is -1

<h3><u>Solution:</u></h3>

Given that when 6 is subtracted from the square of a number, the result is 5 times the number

To find: negative solution

Let "a" be the unknown number

Let us analyse the given sentence

square of a number = a^2

6 is subtracted from the square of a number = a^2 - 6

5 times the number = 5 \times a

<em><u>So we can frame a equation as:</u></em>

6 is subtracted from the square of a number = 5 times the number

a^2 - 6 = 5 \times a\\\\a^2 -6 -5a = 0\\\\a^2 -5a -6 = 0

<em><u>Let us solve the above quadratic equation</u></em>

For a quadratic equation ax^2 + bx + c = 0 where a \neq 0

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here in this problem,

a^2-5 a-6=0 \text { we have } a=1 \text { and } b=-5 \text { and } c=-6

Substituting the values in above quadratic formula, we get

\begin{array}{l}{a=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2 \times 1}} \\\\ {a=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5 \pm \sqrt{49}}{2}} \\\\ {a=\frac{5 \pm 7}{2}}\end{array}

We have two solutions for "a"

\begin{array}{l}{a=\frac{5+7}{2} \text { and } a=\frac{5-7}{2}} \\\\ {a=\frac{12}{2} \text { and } a=\frac{-2}{2}}\end{array}

<h3>a = 6 or a = -1</h3>

We have asked negative solution. So a = -1

Thus the negative solution is -1

6 0
3 years ago
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