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IRISSAK [1]
2 years ago
9

Is 78% a reduction or enlargement

Mathematics
2 answers:
musickatia [10]2 years ago
5 0

Answer:

enlargement

Step-by-step explanation:

<em>hope this helps. i am in algebra two so you can trust my answer. happy holidays and stay safe</em>

<em />

Nikitich [7]2 years ago
5 0
Enlargement


Step by step explanation
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Solve the inequality <br> 52 - 3x &lt; -14
nata0808 [166]

Answer:

x>22

Step-by-step explanation:

Solve the inequality:

-3x<-14-52

-3x<-66

x>22

( You have to switch the sign because you are dividing by a minus)

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3 years ago
Butterfly: wingspan 20 cm silk butterfly: wingspan 4 cm​
MA_775_DIABLO [31]

Answer:

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What is the measure of PRQ?
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Step-by-step explanation:

4 0
3 years ago
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Explain why the polynomial 100x^2 + 150x + 49 is not a perfect square.
Ad libitum [116K]
Answer: The correct answer is Choice C.

For this polynomial to be a perfect square, it would need to be:
(10x + 7)^2

This will ensure that the first terms and the last terms will be 100x^ and 49. However, if you use foil to multiply the factors, you will not get 150x for the center term. Choice C also states that 150x will not be the middle term.
4 0
3 years ago
What is the 9th term of the geometric sequence 4, −20, 100, …?
sveta [45]
First, we are going to find the common ratio of our geometric sequence using the formula: r= \frac{a_{n}}{a_{n-1}}. For our sequence, we can infer that a_{n}=-20 and a_{n-1}=4. So lets replace those values in our formula:
r= \frac{-20}{4}
r=-5

Now that we have the common ratio, lets find the  explicit formula of our sequence. To do that we are going to use the formula: a_{n}=a_{1}*r^{n-1}. We know that a_{1}=4; we also know for our previous calculation that r=-5. So lets replace those values in our formula:
a_{n}=4*(-5)^{n-1}

Finally, to find the 9th therm in our sequence, we just need to replace n with 9 in our explicit formula:
a_{9}=4*(-5)^{9-1}
a_{9}=4*(-5)^{8}
a_{9}=1562500

We can conclude that the 9th term in our geometric sequence is <span>1,562,500</span>
4 0
2 years ago
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