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Rama09 [41]
3 years ago
7

Eric runs a race that can be modeled by the equation shown , where d is his distance, in feet , from the starting line and t is

his time, in seconds, since the race began .
d=8t-4
Which equation shows Eric’s time of distance?

Mathematics
2 answers:
gladu [14]3 years ago
4 0

Answer:D+4/8

Step-by-step explanation:

slamgirl [31]3 years ago
3 0

Answer:

C.  t = (d + 4)/8

Step-by-step explanation:

The equations showing Eric time of distance simply means making t the subject of the formula in the equation given below.

d = 8t - 4

Transfer -4 to the left side and it becomes positive

8t = d + 4

divide both sides by 8

t = (d + 4)/8

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The sum of three numbers is 84, the second number is 2 times the 3rd , the first number is 8 more than the 3rd. What are the num
hjlf

Answer:

a = 27

b = 38

c = 19

Step-by-step explanation:

a + b + c = 84

b = 2c

a = 8+c

8+c +2c+c = 84

4c + 8 = 84

4c = 76

c = 19

b = 2(19) = 38

a = 8 + 19  = 27

5 0
3 years ago
How do I answer this? I'm so confused
mario62 [17]

Answer:

Draw a graph of f(x) and q(x) also find x-intercept and y

8 0
3 years ago
Find the volume of the larger cuboid using given measurement in smaller one.(cu.cm=cubic centimeter
kompoz [17]

Answer:

  c.  12 cu. cm.

Step-by-step explanation:

The larger cuboid is an array of 4 × 3 = 12 of the smaller ones. The volume of the smaller cube is (1 cm)³ = 1 cu. cm. The volume of the larger cuboid is 12 times that value:

  12 × 1 cu. cm = 12 cu. cm.

6 0
3 years ago
8x - 12 = 20 what is x?
PSYCHO15rus [73]

Answer:

4

Step-by-step explanation:

add 12 to 20 then divide 32 by 8

6 0
4 years ago
Read 2 more answers
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

  • dr/dt = + 3 cm/s (positive since it is increasing) and
  • 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

7 0
2 years ago
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