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beks73 [17]
3 years ago
8

Jacks puppy weighed 12 pounds at its first veterinary appointment at the second appointment the puppy's weight had increased by

25% what is the weight in pounds of jacks puppy
Mathematics
1 answer:
Simora [160]3 years ago
3 0

Answer:

15 pounds

Step-by-step explanation:

Given  data

Initial weight= 12 pounds

percent increase in weight = 25%

Let us find the increment

=25/100*12

=0.25*12

=3 pounds

Hence the weight in pounds of jacks puppy is

=3+12

=15 pounds

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The altitude of Death Valley California is 282 ft below sea level of hot air balloon is launched From The Bottom Of Death Valley
asambeis [7]

Answer:

177 ft below sea level.

Step-by-step explanation:

Initial altitude of the hot air balloon = 282 below sea level

Rate of rise of balloon = 10 ft / second

Time = 12 seconds

Total rise in 12 seconds = Rate of rise of balloon \times Time

\Rightarrow 10 \times 12 = 120 ft

Now, we will have to subtract to find the new altitude.

New altitude = 282 - 120 = 162 ft

Now, it is given that balloon Captain lowers the balloon 15 ft.

To find the final altitude, we will have to add it to the previous altitude.

Therefore, the final altitude = 162 + 15 = <em>177 ft below sea level </em>

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3 years ago
Which of the following illustrates the truth value of the given statements? P: 3 is an odd number, and Q: 9 is an odd number (P
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TT - T, since both statements are true, the outcome is true.
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3 years ago
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fomenos
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3 years ago
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Mariana [72]

Answer:

9

Step-by-step explanation:

90 divided by 10 =9

5 0
3 years ago
If the radius of a sphere is increasing at the constant rate of 3 centimeters per second, how fast is the volume changing when t
kotykmax [81]

How fast the volume of the sphere is changing when the surface area is 10 square centimeters is it is increasing at a rate of 30 cm³/s.

To solve the question, we need to know the volume of a sphere

<h3>Volume of a sphere</h3>

The volume of a sphere V = 4πr³/3 where r = radius of sphere.

<h3>How fast the volume of the sphere is changing</h3>

To find the how fast the volume of the sphere is changing, we find rate of change of volume of the sphere. Thus, we differentiate its volume with respect to time.

So, dV/dt = d(4πr³/3)/dt

= d(4πr³/3)/dr × dr/dt

= 4πr²dr/dt where

  • dr/dt = rate of change of radius of sphere and
  • 4πr² = surface area of sphere

Given that

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  • 4πr² = surface area of sphere = 10 cm²,

Substituting the values of the variables into the equation, we have

dV/dt = 4πr²dr/dt

dV/dt = 10 cm² × 3 cm/s

dV/dt = 30 cm³/s

So, how fast the volume of the sphere is changing when the surface area is 10 square centimeters is it is increasing at a rate of 30 cm³/s.

Learn more about how fast volume of sphere is changing here:

brainly.com/question/25814490

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