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schepotkina [342]
3 years ago
8

The sum of a number times 3 and 15. I don't understand it.

Mathematics
1 answer:
Lisa [10]3 years ago
8 0
45.3 would be your anwser multiply 3 and to 1 and then 5 and then 1 add a zero or any numbers that are higher than 9 then put a zero and do the same but the oposite then add it and you will get 45.3 your welcome.
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Complete the patterns of the equivalent ratios by filling in the gaps.
gayaneshka [121]
21:3
9:3
3:1

there all a 3:1 ratio
4 0
3 years ago
Sarah said that 0.00001 is bigger than 0.001 because the first number has more digits to the right of the decimal point. Is Sara
Ivanshal [37]
No, she is wrong. 0.00001 has 4 zeros in front of the one and behind the decimal, making it 1/100000 or 1/10^5. For 0.001, there are 2 zeros in front, making it 1/1000 or 1/10^3. Since a number gets smaller as the denominator gets larger, and 10^5>10^3, 1/10^5<1/10^3
6 0
3 years ago
Answers these two for 5 stars and brainlist and a thanks
shusha [124]

a. Dilation of 2nd equation by a factor of 12

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8 0
3 years ago
Madame Pickney has a rather extensive art collection and the overall value of her collection has been increasing each year. Thre
grigory [225]

Answer:

About $1,178,974

Step-by-step explanation:

Alright, the sequence we have right now is:

500,000       550,000      605,000

We want to find the common ratio, so let's divide 550,000 by 500,000 and see if we get the same value as 605,000 divided by 550,000:

550,000/500,000 = 1.1

605,000/550,000 = 1.1

Now, we know that r = 1.1. We also know the first term is: a_1 = 500,000.

We use the explicit rule of a geometric sequence:

a_n=a_1*r^{n-1}

Here, we want to find a_{10} , which means that n = 10. We know r and a_1, so we just plug in these values:

a_{10}=500,000*(1.1)^{10-1}=500,000*(1.1)^{9} ≈ 1,178,973.85 ≈ 1,178,974

Thus, the answer is about $1,178,974.

Hope this helps!

6 0
3 years ago
Read 2 more answers
If the circle below has a radius of 15cm find each arc length
Elena L [17]
D = 30

I hope this helps you :)
6 0
2 years ago
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