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neonofarm [45]
3 years ago
15

How much greater is 747,784,936 than 373,892,468

Mathematics
2 answers:
svet-max [94.6K]3 years ago
8 0
Difference between these numbers is your answer
747,784,936 - 373,892,468 = 373,892,468
lakkis [162]3 years ago
3 0
The answer is 373892468
You might be interested in
A sphere has a volume of V=900 in cubed. What is the surface area?
ycow [4]
First, we must solve for the radius of the sphere:
V=\frac{4}{3}\pir^{3}
r=(3\frac{V}{4 \pi })^{ \frac{1}{3} }
r=(3*\frac{900}{4 \pi })^{ \frac{1}{3} }
r≈5.99

Second, we must solve for surface area:
A=4\pir^{2}
A=4*\pi*5.99^{2}
A≈450.88 in^{2}
4 0
3 years ago
PLEASE HELP ASAP!!
Sunny_sXe [5.5K]
Y=50+.05x

Just plug in the number of texts for x

y=50+.05(25)
y=51.25

^That was for 25 texts

Now just plug in all the other texts and replace that with x

Hope this helps!
4 0
3 years ago
What is the sum of the first 30 terms of this arithmetic sequence? 6,13,20,27,34
Galina-37 [17]

Answer:

3135

Step-by-step explanation:

Givens

a1 = 6

Use t4 - t3 to get d

t4 = 27

t3 = 20

Step One

Find a1 and d

a1 = 6

d = t4 - t3

d = 27 - 20

d = 7

Step Two

Find the 30th Term

tn= a1 + (n -1 )*d

t30 = 6 + (30 - 1) * 7

t30 = 6 + 29*7

t30 = 6 + 203

t30 = 209

Step Three

Find the sum using Sum = (a + t30)*n/2

n = 30              given

a1 = 6               given  

t30 = 209         calculated from step 2

Sum = (a1 + t30)*n/2        Substitute

Sum = (6+ 209)*30/2       Combine like terms and divide by 2

sum = 215 * 15                  Multiply

Sum = 3135                      Answer

4 0
3 years ago
Help plzzzzzzzzzzzzzzzzz
aleksklad [387]

Answer:99°

Step-by-step explanation:

180-123=57

x=57+42

x=99°

7 0
3 years ago
Read 2 more answers
Find the 14th term in the sequence 1, 1/3, 1/9, … Find the sum of the first 10 terms of the sequence above.
pychu [463]

Answer:

This is a geometric progresion that begins with 1 and each term is 1/3 the preceeding term

   Let Pn represent the nth term in the sequence

 

   Then Pn = (1/3)^n-1

 

   From this P14 = (1/3)^13 = 1/1594323

 

5. The sum of the first n terms of a GP beginning a with ratio r is given by

   Sn = a* (r^n+1 - 1)/(r - 1)

 

   With n = 10, a = 1 and r = 1/3, S10 = ((1/3)^11 - 1)/(1/3 - 1) = 1.500

6 0
3 years ago
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