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3 years ago

How do you find answer for 8k=4/9

Mathematics

2 answers:

nasty-shy [4]3 years ago

8 0

Isolate k by dividing both sides by 8

8k=4/9
k=4/72

Reduce if possible

4/72 -> 1/18

mylen [45]3 years ago

8 0

8k= $\frac{4}{9}$
k = (4÷9)÷8
k = $\frac{4}{9} * \frac{1}{8}$
k = $\frac{4}{72} = \frac{1}{18}$

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Doug works 10.50 per hour working at a restaurant. On Friday he spend 1 3/4 hours cleaning, 2 1/3 paperwork and 1 5/12 hours ser

Lelechka [254]

1 3/4 hours of work equates to $18.375 so $18.38.

2 1/3 hours of work equates to $24.50.

1 5/12 hours of work equates to $14.875 so $14.88

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5 0

2 years ago

Four friends go to an amusement arcade.

galben [10]

Answer:

The rule is:

T(x) = 8x + 20

Explanation:

We are given that:

The rule for the amount that one friend pays is:

C(x) = 2x + 5

Now, we know that:

Each of the four friends will pay the entry fee which is $5 per person

The 4 friends will play the same number of games represented by x

This means that:

We can simply get the rule for the total amount to be paid by the four friends (T(x)) by multiplying the amount paid by each friend by 4

This means that:

T(x) = 4 * C(x)

T(x) = 4(2x + 5)

T(x) = 8x + 20

Hope this helps :)

4 0

3 years ago

.hoi I need a lil help

mihalych1998 [28]

Answer:

Step-by-step explanation:

4+5=9 and 1/2+1/2 =1 so 9+1=10

6 0

2 years ago

Read 2 more answers

Answer this question, thx for the help (the 4th one is this shape O )

RideAnS [48]

Answer:

1st one

Step-by-step explanation:

5 0

2 years ago

You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling

BARSIC [14]

Answer:

(a) 15 m

(b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

d(2) = 25 - (5/2)·4 = 15 . . . . meters

(b) The relative velocity at 2.4 seconds is ...

d'(t) = -5t

d'(2.4) = -5(2.4) = -12.0 . . . m/s

The closure speed with the police car is 12 m/s.

We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters

At the closure speed of 12 m/s, it will take an additional ...

10.6 m/(12 m/s) = 0.8833... s = 53/60 s

to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:

(110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)

= 26 5/36 m/s = 94.1 km/h

_____

Comment on following distance

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

6 0

2 years ago