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2 years ago

Sanjit drives the 45 km distance between York and Leeds he then drives a further 42 km from Leeds to Blackpool

2 answers:

8 0

45 km at 54 km/hr

45 / 54 = 5/6 

It took him 5/6 of an hour, or 50 minute 

61.4117 km/hr

I am Lyosha [343]2 years ago

3 0

Answer: His average speed for his total journey from York to Blackpool is approximately 61.41 km/h.

Step-by-step explanation:

$\text{ The average speed}=\frac{\text{Total distance covered}}{\text{Total time taken}}$

Since, The distance covered by him from York to Leeds = 45 km,

The speed when he covered this distance = 54 km/ h

Thus, the time taken by him in travelling from York to Leeds = 45/54 hours (Because, Time = Distance/speed)

Now, The distance covered by him from Leeds to Blackpool = 42 km,

The time taken by him in travelling from Leeds to Blackpool = 35 minutes = 35/60 hours

Hence, the total time taken by him in this journey

$=\frac{45}{54}+\frac{35}{60}=\frac{450+315}{540}=\frac{765}{540}=\frac{17}{12}$ hours

And, the total distance he covered= 45 + 42 = 87 km

Thus, His average speed

$=\frac{87}{17/12}= \frac{12\times 87}{17}=\frac{1044}{17}=61.4117647\approx 61.41\text{ km per hour}$

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The earth has a mass of approximately 6\cdot 10^{24}6⋅10 24 6, dot, 10, start superscript, 24, end superscript kilograms (\text{

Alex_Xolod [135]

Answer:

0.02

Step-by-step explanation:

The volume of the earth's oceans is approximately 1.34\cdot 10^{9}1.34⋅10

1, point, 34, dot, 10, start superscript, 9, end superscript cubic kilometers (\text{km}^3)(km

)left parenthesis, start text, k, m, end text, cubed, right parenthesis, and ocean water has a mass of about 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\text{km}^3}1.03⋅10

1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start text, k, m, end text, cubed, end fraction .

To simplify, we will use the product of powers property of exponents that says that x^a\cdot x^b = x^{a+b}x

⋅x

a+b

x, start superscript, a, end superscript, dot, x, start superscript, b, end superscript, equals, x, start superscript, a, plus, b, end superscript.

\qquad 1.34\cdot 10^{9}\,\cancel{\text{km}^3} \cdot 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\cancel{\text{km}^3}} = 1.3802 \cdot 10^{21}\,\text{kg}1.34⋅10

⋅1.03⋅10

=1.3802⋅10

kg1, point, 34, dot, 10, start superscript, 9, end superscript, start cancel, start text, k, m, end text, cubed, end cancel, dot, 1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start cancel, start text, k, m, end text, cubed, end cancel, end fraction, equals, 1, point, 3802, dot, 10, start superscript, 21, end superscript, start text, k, g, end text

Hint #2

Next we want to know what portion of the earth's mass this represents. We have:

\qquad \begin{aligned} \dfrac{\text{mass of the oceans}}{\text{total mass of the earth}} &= \dfrac{1.3802 \cdot 10^{21}\,\text{kg}}{6\cdot 10^{24}\,\text{kg}} \\\\ &= \dfrac{1.3802}{6\cdot 10^{3}} \\\\ &= \dfrac{1.3802}{6000} \\\\ &= 0.0002300\overline{3} \end{aligned}

total mass of the earth

mass of the oceans

6⋅10

1.3802⋅10

6⋅10

1.3802

6000

1.3802

=0.0002300

To convert this to a percent, we multiply by 100100100, so the oceans represent 0.02300\overline{3}\%0.02300

%0, point, 02300, start overline, 3, end overline, percent of the earth's total mass, according to these figures.

Hint #3

To the nearest hundredth of a percent, 0.020.020, point, 02 percent of the earth's mass is from oceans.

6 0

2 years ago

Un globo busca aterrizar en medio de dos ciudades A y B, cuya distancia entre si es de 200 km. Si se mide el ángulo de elevación

scoundrel [369]

Answer:

$h \approx 59.898\,km$

Step-by-step explanation:

El diagrama trigonométrico que representa el enunciado es incluido abajo como adjunto. Las siguientes relaciones trigonométricas describen la localización del globo. (The trigonometric diagram representing the statement is included below as attachment. The following trigonometric relations describes the location of the balloon):

$\tan 27^{\circ} = \frac{h}{x}$

$\tan 36^{\circ} = \frac{h}{200-x}$

A continuación, se obtiene la distancia horizontal: (The horizontal distance is obtained hereafter):

$x \cdot \tan 27^{\circ} = (200-x)\cdot \tan 36^{\circ}$

$x \cdot (\tan 27^{\circ} + \tan 36^{\circ})= 200\cdot \tan 36^{\circ}$

$x = \frac{200\cdot \tan 36^{\circ}}{\tan 27^{\circ}+ \tan 36^{\circ}}$

$x \approx 117.557\,km$

La altura aproximada del globo es (The approximated height of the globe is):

$h = x\cdot \tan 27^{\circ}$

$h \approx 59.898\,km$

7 0

2 years ago

Which statement describes function composition with respect to the commutative property? Given f(x) = x² – 4 and g(x) = x – 3, (

Paraphin [41]

Answer:

The correct option are;

f(x) = x² - 4 and g(x) = x - 3, (f ο g)(2) = -3 and (g ο f)(2) = -3, so the function is commutative

Given f(x) = 4·x and g(x) = x², (f ο g)(x) = 4·x² and (g ο f)(x) = 16·x²

So the function is not commutative

Step-by-step explanation:

For the equations f(x) = x² - 4 and g(x) = x - 3, we have;

(f ο g)(x) = f(g(x)) = (x - 3)² - 4 = x² - 6·x + 9 - 4 = x² - 6·x + 5

At x = 2, we have;

(f ο g)(2) = f(g(2)) = 2² - 6×2 + 5 = - 3

Similarly, we have;

(g ο f)(x) = g(f(x)) = x² - 4 -3 = x² - 7

At x = 2, we have;

(g ο f)(2) = g(f(2)) = 2² - 7 = 4 - 7 = -3

Therefore, by commutative property, we have that the result of an operation does not change by changing the order of the operands such that we have;

a + b = b + a or a·b = b·a from which we have resolved also the following operation is commutative

(f ο g)(2) = (g ο f)(2)

Similarly given f(x) = 4·x and g(x) = x², (f ο g)(x) = 4·x² and (g ο f)(x) = 16·x²

So the function is not commutative

(f ο g)(x) = f(g(x)) = 4·x²

(f ο g)(x) = 4·x²

(g ο f)(x) = g(f(x)) = (4·x)² = 16·x²

(g ο f)(x) = 16·x²

∴ (f ο g)(x) = 4·x² ≠ (g ο f)(x) = 16·x²

(f ο g)(x) ≠ (g ο f)(x) the function is not commutative.

8 0

2 years ago

Read 2 more answers

From home, Mary’s work is two thirds along the way to training. Training is 2.5km from work. Mary normally goes to work, then tr

polet [3.4K]

First thing to do is to illustrate the problem, Since it was mentioned that work was along the way to training, the order is shown in the picture. Mary's home and workplace are nearer compared to her training center. It is also mentioned that the distance between work and home, denoted as x, is 2/3 of the total distance from home to training. The total distance is (x + 2.5). Thus,

x = 2/3(x+2.5)
x = 2/3 x + 5/3
1/3 x = 5/3
x = 5 km

Thus, the distance from home to work is 5 km. This means that Mary has to walk this distance twice to return home to get her shoes. Then, she will travel again the total distance of 5+2.5 = 7.5 km to get to her training center. So,

Total distance = 2(5km) + 7.5 km
Total distance = 17.5 km

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2 years ago

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Answer:

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2 years ago