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Pachacha [2.7K]
3 years ago
7

Select ALL numbers that would satisfy the solution to the following inequality

Mathematics
2 answers:
guajiro [1.7K]3 years ago
6 0
The answer is gonna be c
Rashid [163]3 years ago
5 0
I believe the answer should be C
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On a coordinate plane, triangle A B C has points (negative 3, negative 1), (negative 2, negative 3), (1, negative 1) and is refl
andreev551 [17]

Answer:

A' = (-3, 3)

Step-by-step explanation:

If the points are as follows:

A = (-3, -1)

B = (-2, -3)

C = (1, -1)

When reflecting across a y axis the x values remain the same.

When reflecting across y = 1 the y values will remain equal distant from y = 1.

A' = (-3, 3) y value = |-1 - 1| = 2 + 1 = 3

B' = (-2, 5) y value = |-3 - 1| = 4 + 1 = 5

C' = (1, 3) y value = |-1 - 1| = 2 + 1 = 3

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3 years ago
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3 years ago
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Brandi solved 16 / 5 by using a related multiplication expression. What multiplication expression did she is?
klemol [59]

The given question is wrong.

Question:

Brandi solved \frac{1}{6}\div 5 by using a related multiplication expression. What multiplication expression did she is?

Answer:

The related multiplication expression is  $\frac{1}{6}\times \frac{1}{5}.

Solution:

Given expression is \frac{1}{6}\div 5.

We can write 5 as a fraction with denominator 1.

$ \frac{1}{6}\div5=\frac{1}{6}\div \frac{5}{1}

Using the fraction rule  $\frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c},

$\frac{1}{6}\div \frac{5}{1}=\frac{1}{6}\times \frac{1}{5}

Now multiply the fractions using the rule $\frac{a}{b}\times \frac{d}{c}=\frac{a\times d}{b\times c},

$\frac{1}{6}\times \frac{1}{5}=\frac{1\times1}{6\times5}

Let us multiply the numbers in the numerator and denominator separately, we get

$\frac{1\times1}{6\times5}=\frac{1}{30}

Hence the related multiplication expression is  $\frac{1}{6}\times \frac{1}{5}.

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What are numbers that are either whole numbers or negatives of whole numbers
leonid [27]

These are called integer numbers.

We start with natural numbers (or whole numbers), i.e. the set

\mathbb{N} = \{0,1,2,3,4,\ldots\}

When you add the opposite of each whole number, you have the set of integers:

\mathbb{Z} = \{0,\pm1,\pm2,\pm3,\ldots\}

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