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12345 [234]
3 years ago
15

Insert each of the following numbers between two consecutive multiples of 5: 13 37 48 61 93

Mathematics
1 answer:
DedPeter [7]3 years ago
7 0
10 (13) 15
35 (37) 40
45 (48) 50
60 (61) 65
90 (93) 95
You might be interested in
Pls help if you can !
andrey2020 [161]
Length AC
A(-3,-3), C(3,2)

d= (3-(-3))^2 + (2-(-3))^2
d= (6)^2 + (5)^2
d= 36+ 25
Square root 61 = 7.8

Length AC = 7.8


Length BC
B(1,-2) , C(3,2)

d= (3-1)^2 + (2-(-2))^2
d= (2)^2 + (4)^2
d= 4+ 16
Square root 20 = 4.5

Length BC= 4.5

Hope this helps!
7 0
3 years ago
PLEEEEEEAAAAAASSSSSSE HELLLLLP!!
ivolga24 [154]

Answer:The answer is B

Step-by-step explanation:I divided 300 by 5 then added the answer to itself  

300/5=60

60+60=120

7 0
3 years ago
Which is not a common multiple of 4 and 9
Andreas93 [3]

5 I think.

Step-by-step explanation:

4 0
3 years ago
Write the equation for the following relation. Include all of your work in your final answer. \[R=\left\{(x,y):(4,5),\ (8,7),\ (
Alik [6]

Answer:12.9

Step-by-step explanation:

7 0
3 years ago
25 POINTS: In 1960, 21.5% of U.S. households did not have a telephone. This statistic decreased by 75.8% between 1960 and 1990.
dem82 [27]

Answer: 94.8 % (approx)

Step-by-step explanation:

Let the total household in the U.S is x,

⇒ The total household who did not have a telephone in 1960 = 21.5 % of x

= 0.215 x

According to the question,

The total household who do not have a telephone in 1990 = 75.8 % less than the household in 1960 who did not have the cell phone,

= (100-75.8)% of 0.215x

= 24.2 % of 0.215x

= 0.05203x,

⇒ The household in 1990 who had a telephone = x - 0.05203x = 0.94797x

Hence, the percentage of household who has a telephone in 1990

= \frac{0.94797x}{x}\times 100

= 94.797\%\approx 94.8\%

⇒ Around 94.8% household had telephone in the US in 1990.

7 0
4 years ago
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