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3 years ago

Sole equation 4×6-24÷4

Mathematics

1 answer:

Xelga [282]3 years ago

6 0

4x6-24/4

24-24/4

24-6

Answer: 18

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Determine the vertex of the parabola:

Elza [17]

Answer:

Be more clear please

Step-by-step explanation:

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3 years ago

What is the measure of AC?

Firlakuza [10]

Answer:

AC = 52.8

Step-by-step explanation:

33/20 = 1.65

12*1.65 = 19.8

BA = 19.8

CB + BA = AC

33 + 19.8 = 52.8

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2 years ago

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a small rocket is shot from the edge of a cliff suppose that after t seconds the rocket is y meters above the cliff where y=30t-

Natalka [10]

Answer:

Greatest height: 45 meters

Time for greatest height: 3 seconds

Height after 5 seconds: 25 meters above the cliff

Time for height of 40 meters: 7.123 seconds

Height after 7 seconds: -35 meters (35 meters below the cliff)

Step-by-step explanation:

to find the maximum height, we need to calculate the derivative of y in relation to t and then find when dy/dt = 0:

dy/dt = 30 - 10t = 0

10t = 30

t = 3 seconds

In this time, the height is:

y = 30*3 - 5*3^2 = 45 meters

After 5 seconds, the height is:

y = 30*5 - 5*5^2 = 25 meters

The time for the height of 40 meters is:

40 = 30t - 5t^2

t^2 - 6t - 8 = 0

Using Bhaskara's formula, we have:

Delta = 6^2 + 4*8 = 68

sqrt(Delta) = 8.246

t1 = (6 + 8.246) / 2 = 7.123 seconds

t2 = (6 - 8.246) / 2 = -1.123 seconds (negative value for time is not valid)

So the time when the rocket reaches 40 meters is 7.123 seconds

After 7 seconds, the height is:

y = 30*7 - 5*7^2 = -35 meters

The rocket will be 35 meters below the cliff.

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3 years ago

The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m st

Reil [10]

Answer:

a) 30.726m/s and b) 5.5549s

Step-by-step explanation:

a.) What was Chris Huber’s speed in meters per second(m/s)?

Given the distance and time, the formula to obtain the speed is

$v=\frac{d}{t}$ .

Applying this to our problem we have that

$v=\frac{200m}{6.509s}= 30.726m/s$ .

So, Chris Huber’s speed in meters per second(m/s) was 30.726m/s.

b) What was Whittingham’s time through the 200 m?

In a) we stated that $v=\frac{d}{t}$ . This formula implies that

$t=\frac{d}{v}$ .

First, observer that $19\frac{km}{h} =19,000\frac{m}{h}=\frac{19,000}{3,600}m/s= 5.2777m/s$ .

Then, Sam Whittingham speed was equal to Chris Huber’s speed plus 5.2777 m/s. So, $v=30.726\frac{m}{s} +5.2777\frac{m}{s}= 36.003 m/s.$

Then, applying 1) we have that

$t=\frac{200m}{36.003m/s}=5.5549s.$

So, Sam Whittingham’s time through the 200 m was 5.5549s.

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3 years ago

What is the prime factorization of 68

umka21 [38]

68 is not a prime number. The prime factorization of 68 would be 2 x 2 x 17.

6 0

3 years ago

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