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Tresset [83]
2 years ago
15

Algebra help please!!! (Dividing powers)

Mathematics
1 answer:
Sloan [31]2 years ago
5 0
Hi

3^{15}\text{ krill in }3^{7}\text{ m}^3

So  \dfrac{3^{15}}{3^7}=3^{15-7}=3^{8}\text{ krill in 1m}^3
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⚠️⚠️ Help I’m in desperate need of help ⚠️⚠️ the questions are basically in the top picture
Tom [10]

Answer:

1. 3^{3} * 3^{2}

2.6^{-4}

3. 5^{0} < 5^{1}

7 0
2 years ago
How many integers between 10000 and 99999, inclusive, are divisible by 3 or<br> 5 or 7?
Yuki888 [10]

Answer: Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

Step-by-step explanation:

Since we have given that

Integers between 10000 and 99999 = 99999-10000+1=90000

n( divisible by 3) = \dfrac{90000}{3}=30000

n( divisible by 5) = \dfrac{90000}{5}=18000

n( divisible by 7) = \dfrac{90000}{7}=12857.14

n( divisible by 3 and 5) = n(3∩5)=\dfrac{90000}{15}=6000

n( divisible by 5 and 7) = n(5∩7) = \dfrac{90000}{35}=2571.42

n( divisible by 3 and 7) = n(3∩7) = \dfrac{90000}{21}=4285.71

n( divisible by 3,5 and 7) = n(3∩5∩7) = \dfrac{90000}{105}=857.14

As we know the formula,

n(3∪5∪7)=n(3)+n(5)+n(7)-n(3∩5)-n(5∩7)-n(3∩7)+n(3∩5∩7)

=30000+18000+12857.14-6000-2571.42-4258.71+857.14\\\\=48884.15

Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

5 0
3 years ago
How many questions per day i dont even know
nadya68 [22]

Answer:

B. 5( 3n + 2 )...............

6 0
2 years ago
Pleas help will mark brainlist!
Natasha2012 [34]

Geometry

This makes a cube because a cube has 6 faces, 8 vertices, and 12 edges.

8 0
2 years ago
Read 2 more answers
Which of the following is an example of the additive identity?<br> a +(-a) = 0<br> a+0=0<br> a+0= a
UNO [17]

Answer:

a+0= a

Step-by-step explanation:

we know that

The <u>additive identity</u> property says that if you add a real number to zero or add zero to a real number, then you get the same real number back

so

Let

a -----> a real number

a+0=0+a=a

therefore

a+0= a

4 0
2 years ago
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