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Schach [20]
4 years ago
15

Prime Factorization - Prime Factorization how to do it the prime factorization of 1800 help me

Mathematics
2 answers:
hram777 [196]4 years ago
6 0

Answer:

2³ • 3² • 5²

Step-by-step explanation:

aksik [14]4 years ago
3 0

1800|2\\.\ 900|2\\.\ 450|2\\.\ 225|5\\.\ \ 45|5\\.\ \ \ 9|3\\.\ \ \ 3|3\\.\ \ \ 1|\\\\1800=2\cdot2\cdot2\cdot3\cdot3\cdot5\cdot5=2^3\cdot3^2\cdot5^2

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What number has a 9 with a value ten times as many as the 9 in 39,154
Tomtit [17]

Answer:

Any number with 9 in the ten-thousands place. 90,000 is one such number.

Step-by-step explanation:

The 9 in 39,154 is in the thousands place. Its value is 9,000. In order for the 9 in a number to have a value 10 times that, or 90,000, the 9 must be in the ten-thousands place.

There are an infinite number of such numbers. We suspect you have a list you are to choose from. Pick the number with 9 where it is in the number 90,000.

8 0
3 years ago
there were 600 sheep and goats in the hills if the ratio of sheep to goats is 1 to 2 how many are sheep.
larisa86 [58]

Answer:

i think the sheep is 300 because 600 divide by 2 is 300.

5 0
3 years ago
Read 2 more answers
X and y vary inversely, and y =7 when x = 4. What is the constant of variation?
o-na [289]

Answer:

Y doesn't vary directly with x

4 0
4 years ago
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Write an equation for the exponential function represented in the table below.
Neko [114]

Answer:

f(x)=-\frac{1}{2}(10^x)

Step-by-step explanation:

Let the exponential function be f(x)=a(b^x).

When x=0, f(0)=-\frac{1}{2}

-\frac{1}{2}=a(b^0)

a=-\frac{1}{2}

The function now becomes f(x)=-\frac{1}{2}(b^x)

When x=1, f(1)=-5

This implies that:

-5=-\frac{1}{2}(b^1)\\ \implies b=10

Therefore the equation for the exponential function represented in the table below is f(x)=-\frac{1}{2}(10^x)

7 0
4 years ago
Read 2 more answers
Trying to finish this test at 3:30 am plz help
Angelina_Jolie [31]

Given:

The given sum is:

\sum _{k=4}^9(5k+3)

To find:

The expanded form and find the sum.

Solution:

We have,

\sum _{k=4}^9(5k+3)

The expanded form of given sum is:

\sum _{k=4}^9(5k+3)=(5(4)+3)+(5(5)+3)+(5(6)+3)+(5(7)+3)+(5(8)+3)+(5(9)+3)

\sum _{k=4}^9(5k+3)=(20+3)+(25+3)+(30+3)+(35+3)+(4+3)+(45+3)

\sum _{k=4}^9(5k+3)=23+28+33+38+43+48

\sum _{k=4}^9(5k+3)=213

Therefore, the correct option is C.

7 0
3 years ago
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