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Evgen [1.6K]
3 years ago
11

Nine times the difference of a number and 4 is the same as the product of 2 and the same number plus 10. What is the number?

Mathematics
1 answer:
Bezzdna [24]3 years ago
5 0

Answer:

The product of 10 and a number is = 10x Twice that number means 2x 36 less than twice that number= 2x-36. It is given that....The product of 10 and a number is the same as 36 less twice that same number

Step-by-step explanation:

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3 years ago
The weight Wkg of a metal bar varies jointly
maria [59]

Answer:

d = \sqrt{\frac{216W}{35L} }

Step-by-step explanation:

Given that W varies jointly as L and d² then the equation relating them is

W = kLd² ← k is the constant of variation

To find k use the condition W = 140 when d = 4 and L = 54, thus

140 = k × 54 × 4² = 864k ( divide both sides by 864 )

\frac{140}{864} = k , that is

k = \frac{35}{216}

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Multiply both sides by 216

216W = 35Ld² ( divide both sides by 35L )

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Which relation is a function
Morgarella [4.7K]

Answer:

The bottom Right one

Step-by-step explanation:

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3 years ago
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A man travels 20 km by car from Town P to Town Q at an average speed of x km/h. He finds that the time of the journey would be s
yuradex [85]

Answer:

x = 20.

Step-by-step explanation:

First, you should remember the relation:

Distance = Speed*Time.

First, we know that a man travels a distance of 20km at a speed of x km/h, in a time T.

We can write this as:

20km = (x km/h)*T

We know that the time is shortened by 12 minutes if the speed is increased by 5km/h

Rewriting these 12 minutes in hours (remember that 60min = 1 hour)

12 min = (12/60) hours = 0.2 hours

Then from this, he can travel the same distance of 20km in a time T minus 0.2 hours if the speed is increased by 5 km/h

We can write this as:

20km = (x + 5 km/h)*(T - 0.2 h)

Then we have a system of two equations, and we want to find the value of x:

20km = (x km/h)*T

20km = (x + 5 km/h)*(T - 0.2 h)

First, we should isolate the variable T in one of the equations, if we isolate it in the first one, we will get:

20km/(x km/h) = T

Replacing that in the other equation we get:

20km = (x + 5 km/h)*(T - 0.2 h)

20km = (x + 5 km/h)*( 20km/(x km/h) - 0.2 h)

Now we can solve this for x.

Removing the units (that we know that are correct) so the math is easier to read, we get:

20 = (x + 5)*(20/x - 0.2)

We only want to solve this for x.

20 = x*20/x - x*0.2 + 5*20/x - 5*0.2

20 = 20 - 0.2*x + 100/x - 1

subtracting 20 in both sides we get:

20 - 20 = 20 - 0.2*x + 100/x - 1 - 20

0 = -0.2*x + 100/x - 1

If we multiply both sides by x we get:

0 = -0.2*x^2 + 100 - x

-0.2*x^2 - x + 100 = 0

This is just a quadratic equation, we can solve it using the Bhaskara's equation, the solutions are:

x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*(-0.2)*100} }{2*-0.2}  = \frac{1 \pm 9 }{-0.4}

Then the two solutions are:

x = (1 + 9)/-0.4 = -25

x = (1 - 9)/-0.4 = 20

As x is used to represent a speed, the negative solution does not make sense, so we should use the positive one.

x = 20

then the average speed initially is 20 km/h

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Martin still has 30% of the problems left to do.
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