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3 years ago

The scale on a drawing is 1 cm = 4.5 meters. If the length of the drawing is 6.2 cm, what is the actual length in meters?

Mathematics

1 answer:

Nataly [62]3 years ago

7 0

Answer:

I think the awnser is 27.9

Step-by-step explanation:

4.5*6.2=27.9

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Please Help!!!

Komok [63]

Answer:

A = 63 m^2

Step-by-step explanation:

The area of a parallelogram is bh

A = bh

The base is 9 and the height is 7

A = 9*7

A = 63 m^2

3 0

3 years ago

Read 2 more answers

Solve for the x using elimination 3x-2y=13x−2y=1 2x+2y=42x+2y=4

Nikitich [7]

3x - 2y = 1

2x + 2y = 4

Add the second equation to the first

5x = 5

2x + 2y = 4

Divide the first equation by 5

x = 1

2x + 2y = 4

Subtract the first equation from the second

x = 1

x + 2y = 3

Subtract the first equation from the second again

x = 1

2y = 2

Divide the second equation by 2

x = 1

y = 1

<h3>So, the solution is x = 1 and y = 1 {or: (1, 1)} </h3>

7 0

3 years ago

If a cube measures 5.3 cm on each side and has a mass of 280 grams how much is its volume

OlgaM077 [116]

Answer:

8.1 g/cm

Step-by-step explanation:

3 0

3 years ago

Add. Simplify the answer and write it as a mixed number. 1/4+4/5+9/10

djverab [1.8K]

Answer:

39/20 = 1.95

That's my simplified answer

39/20 = 1 19/20

Will this help?

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

3 0

3 years ago