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maks197457 [2]
3 years ago
5

31.5 x _____ = 464 2/4 HELPPPPPPPPPP ASAP

Mathematics
1 answer:
Assoli18 [71]3 years ago
8 0

Answer:

14.7460317

Step-by-step explanation:

you just divide 464.5 by 31.5

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What is the lcd of 7/25 and 4/5
Oliga [24]

Answer:

The answer is 25.

Step-by-step explanation:

This is because 5 goes into 25 and 25 goes into 25 itself. It is the lowest one both denominators fit in. You're welcome.

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What is t=25c=15 plz help
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4 0
3 years ago
Find the value of x.<br> A. 22.5<br> B.40<br> C.45<br> D.95
Troyanec [42]

Answer:

x = 22.5

Step-by-step explanation:

The tangent- secant angle x is half the difference of the measures of the intercepted arcs, that is

\frac{1}{2} (4x + 5 - 50 ) = x ( multiply both sides by 2 )

4x - 45 = 2x ( subtract 2x from both sides )

2x - 45 = 0 ( add 45 to both sides )

2x = 45 ( divide both sides by 2 )

x = 22.5

4 0
3 years ago
When running a 100 meter race Bill reaches his maximum speed when he is 45 meters from the starting line and 5 seconds have elap
Mkey [24]

Bill's speed is constant after 5 seconds and 45 meters into the race.

Correct responses:

a. Bill's maximum speed is 8 m/s

b. 24 meters

c. 57 meters

<h3>Methods used to calculate speed and distance traveled</h3>

Given parameters are;

The distance of the race = 100 meter

Distance at which Bill reaches maximum speed = 45 meters

Speed Bill maintains after 5 seconds = The maximum speed

Time at which Bill is 85 meters from the starting line = 10 seconds after start

a. Required;

Bill's maximum speed in meters.

Solution:

Distance Bill runs at maximum speed, d = 85 m - 45 m = 40 m

Time at which Bill runs the 40 m at maximum speed, <em>t</em> = 10 s - 5 s = 5 s

Speed = \mathbf{\dfrac{Distance}{Time}}

Therefore;

  • Bill's \  maximum \ speed = \dfrac{40 \ m}{5 \ s} = \underline{8 \ m/s}

b. i. Required:

The distance Bill will run for 3 seconds at the maximum speed;

Solution:

Distance,<em> s</em> = Speed, <em>v</em> × Time, <em>t</em>

<em />

The distance traveled at maximum speed in 3 seconds is therefore;

Distance = 8 m/s × 3 s = 24 m

The distance Bill travels in 3 seconds at the maximum speed is 24 meters.

ii) The distance Bill travels at 8 seconds after start, is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 3 seconds = 24 meters

Therefore;

  • Bill's distance from the starting line 8 seconds after start is 45 meters + 25 meters = <u>69 meters</u>

c. Bill's distance 6.5 seconds after the start of the race is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 1.5 seconds = 1.5 s × 8 m/s = 12 meters

  • Bill's distance from the starting line 6.5 seconds after start of the race = 45 meters + 12 meters = <u>57 meters</u>

Learn more about distance, speed, time, relationship here:

https://brainly.in/question/49075584

6 0
2 years ago
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