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kondaur [170]
3 years ago
15

Which inequality is equivalent to this one? y minus 8 less-than-or-equal-to negative 2 y minus 8 + 8 greater-than-or-equal-to ne

gative 2 + 8 y minus 8 + 8 less-than negative 2 + 8 y minus 8 + 2 less-than-or-equal-to negative 2 + 8 y minus 8 + 2 less-than-or-equal-to negative 2 + 2
Mathematics
2 answers:
vodka [1.7K]3 years ago
6 0

Answer:

\boxed{y - 8 + 2\leq  - 2 + 2}

Step-by-step explanation:

Which inequality is equivalent to this one:

y - 8 \leq  - 2

y - 8 + 8\geq  - 2 + 8

y - 8 + 8< - 2 + 8

y - 8 + 2 \leq  - 2 + 8

y - 8 + 2\leq  - 2 + 2

Let’s take the last inequality.

y - 8 + 2\leq  - 2 + 2

Subtract 2 on both sides.

y - 8 + 2-2\leq  - 2 + 2-2

y - 8 \leq  - 2

The inequality is equivalent.

olga2289 [7]3 years ago
5 0

Answer:

d. y minus 8 + 2 less-than-or-equal-to negative 2 + 2

Step-by-step explanation:

Which inequality is equivalent to this one?

<em>y minus 8 less-than-or-equal-to negative 2</em>

y minus 8 + 8 greater-than-or-equal-to negative 2 + 8

y minus 8 + 8 less-than negative 2 + 8

y minus 8 + 2 less-than-or-equal-to negative 2 + 8

y minus 8 + 2 less-than-or-equal-to negative 2 + 2

Take the last option:

y minus 8 + 2 less-than-or-equal-to negative 2 + 2

remove the +2 on each side to get

<em>y minus 8 less-than-or-equal-to negative 2</em>

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55) A dilation of scale factor of 4 will produce a congruent figure<br>True or False​
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Step-by-step explanation:

When a figure is dilated by a scale factor of 4, the new figure produced will be 4 times the size of the original figure. This means that, the size would change while the shape remains the same as the original figure.

Congruent figures must have the same shape and size.

Therefore, the figure produced when a dilation of scale factor of 4 is done is not a congruent figure. Rather, it is a similar figure.

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Read 2 more answers
The sum of the first 10 terms of an arithmetic sequence is 235 and the
Degger [83]

Answer:

             \bold{a_1 = -42.65}\\\\\bold{d=14.7}

Step-by-step explanation:

The sum of the first 10 terms of an arithmetic sequence is:

S_{10}=a_1+a_2+...+a_{10}=\dfrac{a_1+a_{10}}{2}\cdot10=\dfrac{2a_1+(10-1)d}{2}\cdot10

\dfrac{2a_1+(10-1)d}{2}\cdot10=235\\\\(2a_1+9d)\cdot5=235\\\\2a_2+9d=47

the  sum of the second 10 terms is:  a₁₁ + a₁₂+...+ a₂₀

And the sum of the first 20 terms of an arithmetic sequence is:

S_{20}=a_1+a_2+...+a_{10}+a_{11}+...+a_{20}=\dfrac{2a_1+(20-1)d}{2}\cdot10

so the  sum of the second 10 terms is:

a_{11}+a_{12}+...+a_{20}=S_{20}-S_{10}

Therefore we have:

\dfrac{2a_1+(20-1)d}{2}\cdot10-\dfrac{2a_1+(10-1)d}{2}\cdot10=735\\\\(2a_1+19d)\cdot5-(2a_1+9d)\cdot5=735\\\\2a_1+19d-(2a_1+9d)=147\\\\10d=147\\\\d=14.7

and:  

2a_1+9\cdot14.7=47\\\\2a_1+132.3=47\\\\2a_1=-85.3\\\\a_1=-42,65

3 0
3 years ago
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