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saveliy_v [14]
3 years ago
10

A car travels at 865 km in 14 days. At this rate, how far would it travel in 42 days?

Mathematics
2 answers:
Naddika [18.5K]3 years ago
6 0
It would take 2595 days to travel 865 km in 14 days
GarryVolchara [31]3 years ago
3 0
14 times 3 equals 42 and 865 times 3 equals 2595
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4. A rectangular field measures, 616m by 456m. Fencing
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Answer:

The number of posts required to fence the field are;

136 posts

Step-by-step explanation:

The measure of the rectangular field is given as follows;

The length of the rectangular field = 616 m

The width of the rectangular field = 456 m

Let 'x' represent the 'furthest distance possible between the posts

There will be one post at each of the four corners of the field

Let 'c' represent the number of posts at the four corners of the field

Therefore, the number of posts at the corner of the field, c = 4

Therefore, 'x' is highest common factor of 616 and 456

The prime factorization of 616 = 2³ × 7¹ × 11¹

The prime factorization of 456 = 2³ × 9 × 19

Therefore, the highest common factor of 616 and 456 = 2³ = 8

The distance between posts = 8 m

Let 'w' represent the number of posts on the 456 m side of the field  less the posts at the corner corner of the field and let 'l' represent the number of posts on the 616 m side of the field  less the posts at the corner corner of the field, we have;

The number of posts on the 456 m side less the corner post, w = 456/8 - 1 = 57 - 1 = 56

The number of posts on the 616 m side less the corner post, l = 616/8 - 1 = 77 - 1 = 76

The total number of posts required to fence the field, T = w + l + c

∴ T = 56 + 76 + 4 = 136

The total number of posts required to fence the field, T = 136 posts

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lutik1710 [3]

Answer:

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Step-by-step explanation:

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3 years ago
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T fewer than the product of 9 and 8
ale4655 [162]

Answer: t

Step-by-step explanation:

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How many solutions does a system of equations have when solving results in the statement 3=5?
Tems11 [23]

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Step-by-step explanation:

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So no solution

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What is the square root of 28x to the third power y to the third power
Colt1911 [192]

Hello!

√28x³y³ is equal to 2xy\sqrt{7xy}

We can find this solution through the following steps.

  1. Rewriting the equation by factoring 4 from 28, factoring out x², and factoring out y²
  2. We then can rewrite 4 as 2² in our equation, leaving us with \sqrt{2^{2}*7*(x^{2}x)*y^{2}y }
  3. We will then move our x and 7, and rewrite 2^{2}*x^{2}* y^{2} as (2xy)^{2}
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