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Katen [24]
3 years ago
8

HELP ME

Mathematics
1 answer:
navik [9.2K]3 years ago
7 0
$97.20 divided by $9.25 is 10 hours
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Harriet has 3/4tanks of gas left in her car .if she needs 1/4 tank to go to her friends house and another 1/4 tank to get back h
givi [52]
Yes, because 1/4 plus 1/4 equals 2/4 (or 1/2) and 3/4-2/4 equals 1/4.

7 0
3 years ago
4.95 what value of digit is the 9
belka [17]

Note:

4 is in the ones place

. (decimal)

9 is in the tenths place

5 is in the hundredths place

tenths place has the value digit of 9

hope this helps

4 0
3 years ago
What is AE ? <br><br><br> ___ units
sattari [20]


We can form two ratios to help us solve this:

(2x+6):10 = (x+6):8


(2x+6)         (x+6)
---------   =   --------
   10              8


Cross multiply:
8(2x+6) = 10(x+6)

Foil:
16x + 48 = 10x + 60

Move like variables to the same side:
16x - 10x = 60 - 48
6x = 12
x = 2

Line AE = 2x+6
Plug 2 for x into the equation: 2(2) + 6 = 10

Hope this helps, and May the Force Be With You!

-Jabba



6 0
3 years ago
In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates
Anettt [7]

Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

S = ut + \frac{1}{2} at^{2}

Where S is distance

u is the initial velocity

a is acceleration

and t is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

S = S₁

u = 0 m/s ( The player starts at rest)

a = 2.2 m/s²

Then,

S₁ = 0(t) + \frac{1}{2} (2.2)t^{2}

S₁ = 1.1t^{2}

t^{2} = \frac{S_{1} }{1.1}

For player 2

S = S₂

u = 0 m/s ( The player starts at rest)

a = 1.7 m/s²

Then,

S₂ = 0(t) + \frac{1}{2} (1.7)t^{2}

S₂ = 0.85t^{2}

t^{2} = \frac{S_{2} }{0.85}

Since the time spent by both players is equal, We can  write that

t^{2} = t^{2}

Then,

\frac{S_{1} }{1.1} =  \frac{S_{2} }{0.85}

0.85S_{1} = 1.1S_{2}

From, S₁ + S₂ = 41 m

S₂ = 41  - S₁

Then,

0.85S_{1} = 1.1 ( 41 -S_{1} )

0.85S_{1} = 45.1 - 1.1S_{1}

0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1}  = 23.13 m

This is the distance covered by the first player. We can then put this value into t^{2} = \frac{S_{1} }{1.1} to determine how much time elapses before the players collide.

t^{2} = \frac{S_{1} }{1.1}

t^{2} = \frac{23.13 }{1.1}

t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs

Hence, the time that elapses before the players collide is 4.59 secs

3 0
4 years ago
944-6552-9711 psw zV3WZ3<br><br> bo y ss
NemiM [27]

Answer:

Uhm okay

Step-by-step explanation:

Wanna talk I’m bored :/

3 0
3 years ago
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