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velikii [3]
3 years ago
15

Please help thank u!

Mathematics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

6=65

Step-by-step explanation:

34

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A baseball fits snugly inside a transparent display cube. The length of an edge of the cube is 2.9 inches.
kozerog [31]

Answer:

it would be just a little less to fit in

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Write the point-slope form of the line that passes through (5, 5) and is PARALLEL to a line with a slope of 1/4. Include all of
lesya692 [45]
Parallel lines will have the same slope

y - y1 = m(x - 1)
slope(m) = 1/4
(5,5)...x1 = 5 and y1 = 5
now we sub
y - 5 = 1/4(x - 5) <==== ur parallel line
==================
perpendiculr lines will have a negative reciprocal slope. To find the negative reciprocal, u flip the number and change the sign. So the negative reciprocal of 1/4 is -4....see how I flipped 1/4 and made it 4/1...then changed the sign making it -4/1 or just -4. That will be ur slope of the perpendicular line.

y - y1 = m(x - x1)
slope(m) = -4
(5,5)...x1 = 5 and y1 = 5
now we sub
y - 5 = -4(x - 5) <=== ur perpendicular line
4 0
3 years ago
25x89 divide by 12 then times it by 9
Soloha48 [4]

Answer:

1668.75

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
.Given the functions f(n) = 500 and g(n) = (nine tenths)n _ 1, combine them to create a geometric sequence, an, and solve for th
weeeeeb [17]
The original functions are: f(n) = 500 and g(n) = [9/10]^(n-1)

A geometric sequence combining them is: An = f(n)*g(n) = 500*[9/10]^(n-1):

Some terms are:
A1= 500
A2 = 500*[9/10]
A3 = 500*[9/10]^2
A4 = 500*[9/10]^3
....
A11 = 500*[9/10]^10 ≈ 174.339

Answer: the third option, An = 500[9/10]^(n-1); A11 = 174.339

 
4 0
3 years ago
Read 2 more answers
What is a three digit number thats an odd multiple of three and the product of its digits is 24 and is larger than 15 squared?
Reika [66]
The only way 3 digits can have product 24 is 
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4 
To be divisible by 3 the sum of the digits must be divisible by 3.
1+  3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3. 

So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.

Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.

The others must end in 3. 

They must be greater than 152 which is 225. So the

First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.

The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.  
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
7 0
3 years ago
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