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AURORKA [14]
3 years ago
13

The amount in a savings account increased from 200 to 216. What was the percent of increase?

Mathematics
1 answer:
aleksandrvk [35]3 years ago
4 0
216/200=%
1.08 which equals 108%
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4. Ashley claims that if she triples the side length of a square (s), the area of the square will also be tripled. Do you agree?
stepladder [879]

<u>Answer</u><u> </u><u>:</u><u>-</u>

Let the side of the square measure a cm, thus, according to the question,

  • New length = 3(a cm)

We know that,

➭ Area ( Square ) = a²

Thus, the area of the square with normal length i.e. a cm, will be,

⤳ Area = a x a

⤳ Area = a²

Now, by taking the new length, the area will become,

⤳Area = 3a x 3a

⤳ Area = 9a²

Now,

We observe that, if we divide these, we get,

➤ 9a²/a²

( a² will get canceled since its common )

Thus, the ratio becomes,

➭ 9 : 1 = Area square with tripled length : Area of square

Thus, the area will increase by 9 times, and not 3.

Hence, Ashley's assumptions are wrong.

5 0
2 years ago
Simplify this expression. 6^7 ÷ 6^5 A. 30 B. 35 C. 36 D. 42
bulgar [2K]

Answer:

C 36

Step-by-step explanation:

When we have to exponents to the same base, and we are dividing them, we can subtract the exponents

a^b ÷ a^c = a^ (b-c)

6^7 ÷ 6^5 = 6^(7-5)

                 = 6^(2)

                = 36

6 0
3 years ago
Read 2 more answers
How was life different for Romans with low incomes and wealthy Romans? Romans living in poverty had few duties and more free tim
ahrayia [7]

Wealthy Romans relied on servants to run their households.

brainly.com/question/16395373

8 0
2 years ago
If f(p) divided by x-p and x-q have the same remainder
oee [108]
Hello,


x^2-y^2=(x+y)(x-y)
x^3-y^3=(x-y)(x²+xy+y²)

Let's use Horner's division

.........|a^3|a^2.|a^1..........|a^0
.........|1....|5....|6..............|8....
a=p...|......|p....|5p+p^2....|6p+5p^2+p^3
----------------------------------------------------------
.........|1....|5+p|6+5p+p^2|8+6p+5p^2+p^3

The remainder is 8+6p+5p^2+p^3 or 8+6q+5q^2+q^3


Thus:
8+6p+5p^2+p^3 = 8+6q+5q^2+q^3
==>p^3-q^3+5p^2-5q^2+6p-6p=0
==>(p-q)(p²+pq+q²)+5(p-q)(p+q)+6(p-q)=0
==>(p-q)[p²+pq+q²+5p+5q+6]=0 or p≠q
==>p²+pq+q²+5p+5q+6=0

And here, Mehek are there sufficients explanations?
3 0
3 years ago
I need help in the question 7
Alona [7]
6.75 in by 5 in. the area is 33.75
4 0
3 years ago
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