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2 years ago

a car travels 20 meters in half a minute. if a car travels 10 meters, how many seconds will that take? 1 minute= 60 seconds

Mathematics

2 answers:

snow_tiger [21]2 years ago

5 0

Answer:

20 meters= 30 seconds

10 meters = x seconds

10*30=40/20

=15

Step-by-step explanation:

it will take 15 seconds good luck

butalik [34]2 years ago

4 0

Answer:

15 seconds

Step-by-step explanation:

If it travels 20 meters in 30 seconds that means it travels 10 meters in 15 seconds

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Step-by-step explanation:

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3 years ago

Select all the equations that are true when x = 4

V125BC [204]

Answer:

B and E

Step-by-step explanation:

A. 8x = 2

8(4) = 2

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B. 19 + x = 23

19 + 4 = 23

23 = 23

C. 40/x = 5

40/4 = 5

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D. -3x = 12

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7 0

3 years ago

Please helppppp meeeeeee!!!!!!!!!!!!!!!!!!!!

astraxan [27]

Answer:option A is the correct answer.

Step-by-step explanation:

The given triangle UVW is a right angle triangle. This is because one of its angles is 90 degrees. The sum of the two other angles is 180. To determine Cos U, we will apply trigonometric ratio.

Cos U = adjacent side/ hypotenuse.

Looking at the triangle with C as the reference angle,

Adjacent side = 12

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6 0

3 years ago

The manufacturer of a CD player has found that the revenue R (in dollars) is Upper R (p )equals negative 5 p squared plus 1 com

AleksAgata [21]

Answer:

The maximum revenue is $1,20,125 that occurs when the unit price is $155.

Step-by-step explanation:

The revenue function is given as:

$R(p) = -5p^2 + 1550p$

where p is unit price in dollars.

First, we differentiate R(p) with respect to p, to get,

$\dfrac{d(R(p))}{dp} = \dfrac{d(-5p^2 + 1550p)}{dp} = -10p + 1550$

Equating the first derivative to zero, we get,

$\dfrac{d(R(p))}{dp} = 0\\\\-10p + 1550 = 0\\\\p = \dfrac{-1550}{-10} = 155$

Again differentiation R(p), with respect to p, we get,

$\dfrac{d^2(R(p))}{dp^2} = -10$

At p = 155

$\dfrac{d^2(R(p))}{dp^2} < 0$

Thus by double derivative test, maxima occurs at p = 155 for R(p).

Thus, maximum revenue occurs when p = $155.

Maximum revenue

$R(155) = -5(155)^2 + 1550(155) = 120125$

Thus, maximum revenue is $120125 that occurs when the unit price is $155.

6 0

3 years ago