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3 years ago

A train travels at a speed of 48.6 km/h. Express this speed in(ii) cm/min.(1) m/s,

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

4 0

Answer:

Step-by-step explanation:

Conversion factors:

1 km = 1000 m

1 m = 100 cm

1 h = 60 min

1 min = 60 sec

48.6 km/h = 81,000 cm/min = 13.5 m/sec

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Apr 17, 1:48:02 AM

galben [10]

Answer:

y = 9.54

and, x = 0.08

Step-by-step explanation:

Given that

x - 2y = -19 .......................(1)

3x + 5y = 48 ...............(2)

Based on the above information

Multiply by 3 in equation 1

So,

3x - 6y = -57

3x + 5y = 48

- - -

-11y = -105

y = 9.54

So, the x would be

x - 2(9.54) = -19

x - 19.08 = -19

x = -19 + 19.08

x = 0.08

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2 years ago

The figure below is the graph of the dimensions of a rectangle whose adjacent side lengths exhibit inverse variation.

cricket20 [7]

Answer:

The answer is true just answered

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3 years ago

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What is the surface area of a cylindrical ring where the diameter of the cross section is 6.3 in and the center line has a lengt

Kruka [31]

The surface area of the cylindrical ring is given by
πdh
where d is the diameter and h is the height:
π•6.3•48 = 950.02
The answer is C. 950.02 in^2

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3 years ago

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Chnstopher earned $142.20. He saved $24.50. He spent the rest on 2 new pairs

pogonyaev

Answer:

$58.25

Step-by-step explanation:

142.20 - 24.50 = 117.50

117.50 / 2 = 58.25

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3 years ago

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What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

2 years ago

A train travels at a speed of 48.6 km/h. Express this speed in(ii) cm/min.(1) m/s,​

A train travels at a speed of 48.6 km/h. Express this speed in(ii) cm/min.(1) m/s,