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denis23 [38]
3 years ago
12

Jawad has seven rocks.

Mathematics
1 answer:
ad-work [718]3 years ago
3 0

680 grams.

So, it's telling us to calculate the weight, in grams, of the other rock.

So firstly,

6 of the rocks EACH have a weight of 720 grams.

So, 6 x 720 = 4320

Now save 4320 for later.

Next, we need to convert the 5 kilograms into grams.

So, 5 x 1000 = 5000

now do, 5000 - 4320 = 680.

Your final answer would then be = 680g

I hope this helped.

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Jimmy is going scuba diving. He started at 10 ft below sea level. He then dove 15 ft further, then he got nervous and came up 5
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Answer:

it is 55

Step-by-step explanation:

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Given that r is the midpoint of qs and that the lenght of qr is 5.7, find the lenght of qs
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We know that the midpoint is the a point in the center of the line (r). This means that from the midpoint to the endpoint (qr) is half the length of the overall line. This means that 5.7 (the length of qr) units is half the length of the line. Two halves make a whole so 5.7 × 2 = 11.4 units as the length of the entire line qs.
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If the 5th term of a geometric progression is 162 and the 8th term is 4374, find the (i) 1st three terms of the sequence; (ii) s
Alona [7]

Answer:

see explanation

Step-by-step explanation:

The nth term of a geometric progression is

a_{n} = a₁r^{n-1}

where a₁  is the first term and r the common ratio

Given a₅ = 162 and a₈ = 4374 , then

a₁r^{4} = 162 → (1)

a₁r^{7} = 4374 → (2)

Divide (2) by (1)

\frac{a_{1}r^{7}  }{a_{1}r^{4}  } = \frac{4374}{162}

r³ = 27

r = \sqrt[3]{27} = 3

Substitute r = 3 into (1) and solve for a₁

a₁(3)^{4} = 162

81a₁ = 162

a₁ = \frac{162}{81} = 2

Then

a₂ = a₁ × 3 = 2 × 3 = 6

a₃ = a₂ × 3 = 6 × 3 = 18

The first 3 terms are 2, 6, 18

(ii)

The sum to n terms of a geometric progression is

S_{n} = \frac{a_{1}(r^{n}-1)  }{r-1} , then

S_{10} = \frac{2(3^{10}-1) }{3-1}

     = \frac{2(59049-1)}{2}

     = 59049 - 1

     = 59048

   

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