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mixas84 [53]
3 years ago
7

James bike 5.40 miles Fred bike 9.2 how much farther did James ride then Fred in a fraction

Mathematics
1 answer:
olga55 [171]3 years ago
6 0
3 8/10 of a mile. You subtract 5.4 from 9.2 and get 3.8 and 3.8 in fraction form is 3 and 8/10. 3 8/10 simplyfied is 3 4/5
You might be interested in
Does the addition problem show a way to add 27 + 38?
Dmitry [639]

Answer:

a. yes

b. no

c. yes

d. yes

Step-by-step explanation:

For a, 27 + 38 can be broken apart. 27 is broken up by adding smaller numbers (20 + 7 = 27) and the same is done with 38 (30 + 8 = 38), so A and C shows a way to add 27 and 38. In C, the numbers are just put into a different order. B is not a way to add 27 and 38, because the sum is different.

27 + 38 = 65, however 20 + 70 + 38 = 128. The addition problem for D is a way to solve for 27 + 38, because it is broken up differently than A and C. They instead added the 20 and 30 together first, then split up 15 (from 8+7) into 10 and 5. So D is a way to solve, because it gets the same answer as 27 and 38 :D

8 0
3 years ago
This the questions that have me stuck​
IRINA_888 [86]
First a and then b.
4 0
3 years ago
What is (6.89+14.52)+(-14.52)???
MrRa [10]
The answer is 6.89 because
6.89+14.52=21.41
So 21.41+(-14.52)=6.89
7 0
3 years ago
Raymond uses the Venmo
Arisa [49]

Answer:

2.29

Step-by-step explanation:

2.95 plus 1.26 is 4.21, and 6.50 minus 4.21 is 2.29.

5 0
3 years ago
Express the terms of the following geometric sequence recursively.
BabaBlast [244]

Answer:

The most correct option for the recursive expression of the geometric sequence is;

4. t₁ = 7 and tₙ = 2·tₙ₋₁, for n > 2

Step-by-step explanation:

The general form for the nth term of a geometric sequence, aₙ is given as follows;

aₙ = a₁·r⁽ⁿ⁻¹⁾

Where;

a₁ = The first term

r = The common ratio

n = The number of terms

The given geometric sequence is 7, 14, 28, 56, 112

The common ratio, r = 14/7 = 25/14 = 56/58 = 112/56 = 2

r = 2

Let, 't₁', represent the first term of the geometric sequence

Therefore, the nth term of the geometric sequence is presented as follows;

tₙ = t₁·r⁽ⁿ⁻¹⁾ = t₁·2⁽ⁿ⁻¹⁾

tₙ =  t₁·2⁽ⁿ⁻¹⁾ = 2·t₁2⁽ⁿ⁻²⁾ = 2·tₙ₋₁

∴ tₙ = 2·tₙ₋₁, for n ≥ 2

Therefore, we have;

t₁ = 7 and tₙ = 2·tₙ₋₁, for n ≥ 2.

4 0
3 years ago
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