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Margarita [4]
3 years ago
13

A racing car used 255 litres of fuel to complete a 340 km race. On average, how many litres of fuel did the care use every 100 k

m?
Mathematics
1 answer:
zhenek [66]3 years ago
3 0
The answer is 75 liters per 100km. If you divide 340 by 100 you get 3.4 and then divide 255 by 3.4 you get 75.
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Write an equation for each sentence 13 times the sum of a number and 5<br> is 91<br> I
Volgvan

Answer:13( n + 5) = 91

Step-by-step explanation: solve the parenthesis first and then multiply that by 13

7 0
3 years ago
Answer correctly for brainliest pleaseeee!!
wolverine [178]

Answer:

A. (x,y) → (-x,y)

Step-by-step explanation:

A: (-5,-1)
A': (5,1)
B: (-3, -3)
B': (3, -3)
C: (-5, -5)

C': (5, -5)
D: (-7, -3)

D': (7,-3)

The x values change sign but the y values don't.

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2 years ago
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What is the volume of the right rectangular prism
11111nata11111 [884]
I don’t get this sorry
8 0
3 years ago
the length of a rectangle is 5ft more than twice the width and teh area of the rectabgle is 42ft squared. what are the dimension
lara [203]
The width is 3.5 and the length is 12

4 0
3 years ago
72
Zigmanuir [339]

Answer:

Ai. Arithmetic sequence

Aii. Tn = 5 + 7n

Bi. Geometric

Bii. Tn = 8 × 2ⁿ¯¹

Step-by-step explanation:

To successfully answer the questions given above, note the following:

1. If the sequence is Arithmetic, then:

2nd – 1st = 3rd – 2nd = common difference (d)

2. If the sequence is geometric, then,

2nd / 1st = 3rd / 2nd = common ratio (r)

3. A sequence can not be arithmetic geometric at the same time.

4. The nth term of arithmetic sequence is:

Tn = a + (n – 1)d

5. The nth term of geometric sequence is:

Tn = arⁿ¯¹

A. Sequence => 12, 19, 26

i. Determination of the type of sequence.

We'll begin by calculating the common difference

1st term = 12

2nd term = 19

3rd term = 26

Common difference (d) = 2nd – 1st

d = 19 – 12 = 7

OR

d = 3rd – 2nd

d = 26 – 19 = 7

Since a common difference exist in the sequence, the sequence is arithmetic sequence.

ii. Determination of the nth term.

Common difference (d) = 7

1st term (a) = 12

nth term (Tn) =?

Tn = a + (n – 1)d

Tn = 12 + (n – 1)7

Tn = 12 + 7n – 7

Tn = 5 + 7n

B. Sequence => 8, 16, 32

Bi. Determination of the type of sequence.

Let us begin by calculating the common ratio.

1st term = 8

2nd term = 16

3rd term = 32

Common ratio (r) = 2nd / 1st

r = 16 / 8

r = 2

OR

r = 3rd / 2nd

r = 32 / 16

r = 2

Since a common ratio exist in the sequence, the sequence is geometric.

Bii. Determination of the nth term.

Common ratio(r) = 2

1st term (a) = 8

nth term =?

Tn = arⁿ¯¹

Tn = 8 × 2ⁿ¯¹

8 0
2 years ago
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