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olganol [36]
3 years ago
7

I NEED HELP ASAP IM TIMED... picture connected!

Mathematics
1 answer:
cluponka [151]3 years ago
5 0

Answer:

d

Step-by-step explanation:

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Batteries come in packs of 4. Last month, Joe used 58 batteries in his remote control drone. How many packs of batteries did he
vovangra [49]

It is given that batteries come in a packs of 4. It means in each pack there are 4 batteries.

Joe used 58 batteries . So to find the total number of packets of batteries joe has to open is

Number of batteries used / Total number of batteries in each packet

= 58 / 4

= 14.5

The number of battery can not be in decimal. So we will round the answer to integer. If we round it to 14 it means 14 packets. But in 14 packets there are 14*4 = 56 batteries .

But we know that Joe used 58 batteries. So we will round the final answer to 15.

It means Joe has to open 15 packets of batteries.

8 0
3 years ago
What is 1/3 of 2/3?​
Alla [95]
1/3 of 2/3

Just multiply them

1/3 x 2/3 = 2/9

5 0
3 years ago
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A man travels 20 km by car from Town P to Town Q at an average speed of x km/h. He finds that the time of the journey would be s
yuradex [85]

Answer:

x = 20.

Step-by-step explanation:

First, you should remember the relation:

Distance = Speed*Time.

First, we know that a man travels a distance of 20km at a speed of x km/h, in a time T.

We can write this as:

20km = (x km/h)*T

We know that the time is shortened by 12 minutes if the speed is increased by 5km/h

Rewriting these 12 minutes in hours (remember that 60min = 1 hour)

12 min = (12/60) hours = 0.2 hours

Then from this, he can travel the same distance of 20km in a time T minus 0.2 hours if the speed is increased by 5 km/h

We can write this as:

20km = (x + 5 km/h)*(T - 0.2 h)

Then we have a system of two equations, and we want to find the value of x:

20km = (x km/h)*T

20km = (x + 5 km/h)*(T - 0.2 h)

First, we should isolate the variable T in one of the equations, if we isolate it in the first one, we will get:

20km/(x km/h) = T

Replacing that in the other equation we get:

20km = (x + 5 km/h)*(T - 0.2 h)

20km = (x + 5 km/h)*( 20km/(x km/h) - 0.2 h)

Now we can solve this for x.

Removing the units (that we know that are correct) so the math is easier to read, we get:

20 = (x + 5)*(20/x - 0.2)

We only want to solve this for x.

20 = x*20/x - x*0.2 + 5*20/x - 5*0.2

20 = 20 - 0.2*x + 100/x - 1

subtracting 20 in both sides we get:

20 - 20 = 20 - 0.2*x + 100/x - 1 - 20

0 = -0.2*x + 100/x - 1

If we multiply both sides by x we get:

0 = -0.2*x^2 + 100 - x

-0.2*x^2 - x + 100 = 0

This is just a quadratic equation, we can solve it using the Bhaskara's equation, the solutions are:

x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*(-0.2)*100} }{2*-0.2}  = \frac{1 \pm 9 }{-0.4}

Then the two solutions are:

x = (1 + 9)/-0.4 = -25

x = (1 - 9)/-0.4 = 20

As x is used to represent a speed, the negative solution does not make sense, so we should use the positive one.

x = 20

then the average speed initially is 20 km/h

3 0
3 years ago
What is this answer?
zhuklara [117]

Answer:

B is the correct answer

3 0
3 years ago
Suppose y varies directly with x. If y = –4 when x = 8, what is the equation of direct variation? Complete the steps to write th
rosijanka [135]
Direct variation is of the form y=kx.  We are given the point (8, -4) so we can solve for k, the constant of variation...

-4=8k, divide both sides by 8

-4/8=k

k=-0.5, so our equation is:

y=-0.5x, 
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