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Dennis_Churaev [7]
3 years ago
11

Pratibha spends 5 3/4 hours in school. He spends 4 1/4 hours for classes and 3/4 hours for lunch break. She spends the remaining

time on sports. How much time does she spend on sports?
Mathematics
1 answer:
zzz [600]3 years ago
7 0

Answer:

13 hours and 15 minutes of her day are spent on sports if she doesn't sleep.

Step-by-step explanation:

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A function which has a constant difference per interval is?
QveST [7]

Answer:

I think A or C I'm not sure tho

5 0
3 years ago
PLZ HURRY IT'S URGENT!
Furkat [3]

Answer:

D. 12 units

Step-by-step explanation:

The distance between two points is found by

d = sqrt( (x2-x1)^2 + (y2-y1)^2)

   sqrt( (4 -4)^2 + (-5-7)^2)

 sqrt((0^2 + (-12)^2)

 sqrt(0+144)

sqrt(144)

12

4 0
3 years ago
5 1/4 written in radical form
emmainna [20.7K]

Answer:Exact Form:

4 √ 5

Step-by-step explanation:

5 0
3 years ago
Javier's fuel tank holds 12 3/4 gallons of gasoline when completely full. He had some gas in the tank and added 10.3 gallons of
GenaCL600 [577]

Answer:

2.45 gallons of gasoline were in the tank before.

Step-by-step explanation:

Consider the provided information.

The tank can hold 12\frac{3}{4} gallons of gasoline when completely full. He had some gas in the tank and added 10.3 gallons of gasoline to fill it completely.

We need to find: How many gallons of gasoline were in the tank before Javier added some?

For this, we need to subtract added gallons of gasoline from the capacity of the tank.

12\frac{3}{4}-10.3=\frac{51}{4}-10.3

=12.75-10.3

=2.45

Hence, 2.45 gallons of gasoline were in the tank before.

6 0
3 years ago
Given f'(x)=(x-4)(6-2x) find the x-coordinate for the relative minimum of the graph f(x).
vlabodo [156]
<span>The graph you plotted is the graph of f ' (x) and NOT f(x) itself. </span>

Draw a number line. On the number line plot x = 3 and x = 4. These values make f ' (x) equal to zero. Pick a value to the left of x = 3, say x = 0. Plug in x = 0 into the derivative function to get

f ' (x) = (x-4)(6-2x)
f ' (0) = (0-4)(6-2*0)
f ' (0) = -24

So the function is decreasing on the interval to the left of x = 3. Now plug in a value between 3 and 4, say x = 3.5

<span>f ' (x) = (x-4)(6-2x)
</span><span>f ' (3.5) = (3.5-4)(6-2*3.5)
</span>f ' (3.5) = 0.5

The function is increasing on the interval 3 < x < 4. The junction where it changes from decreasing to increasing is at x = 3. This is where the min happens.

So the final answer is C) 3
5 0
3 years ago
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