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padilas [110]
3 years ago
12

Ian wants to run 412 miles to get ready for his upcoming race.

Mathematics
2 answers:
tankabanditka [31]3 years ago
5 0

Answer:

137.3 miles apart from each other

Step-by-step explanation:


ivanzaharov [21]3 years ago
4 0

Answer:

3/2

Step-by-step explanation:

i jus did the quiz and got 100% . your welcome

You might be interested in
Someeee one?????????????????
Ad libitum [116K]

Answer:

Option B) a_{n} = 2\cdot 4^{n-1}

Step-by-step explanation:

The given geometric sequence is

2, 8, 32, 128,....

The general form of a geometric sequence is given by

a_{n} = a_{1}\cdot r^{n-1}

Where n is the nth term that we want to find out.

a₁ is the first term in the geometric sequence that is 2

r is the common ratio and can found by simply dividing any two consecutive numbers in the sequence,

r=\frac{8}{2} = 4

You can try other consecutive numbers too, you will get the same common ratio

r=\frac{32}{8} = 4

r=\frac{128}{32} = 4

So the common ratio is 4 in this case.

Substitute the value of a₁ and r into the above general equation

a_{n} = 2\cdot 4^{n-1}

This is the general form of the given geometric sequence.

Therefore, the correct option is B

Note: Don't multiply the first term and common ratio otherwise you wont get correct results.

Verification:

a_{n} = 2\cdot 4^{n-1}

Lets find out the 2nd term

Substitute n = 2

a_{2} = 2\cdot 4^{2-1} = 2\cdot 4^{1} = 2\cdot 4 = 8

Lets find out the 3rd term

Substitute n = 3

a_{3} = 2\cdot 4^{3-1} = 2\cdot 4^{2} = 2\cdot 16 = 32

Lets find out the 4th term

Substitute n = 4

a_{4} = 2\cdot 4^{4-1} = 2\cdot 4^{3} = 2\cdot 64 = 128

Lets find out the 5th term

Substitute n = 5

a_{5} = 2\cdot 4^{5-1} = 2\cdot 4^{4} = 2\cdot 256 = 512

Hence, we are getting correct results!

6 0
3 years ago
Sherri saves nickels and dimes in a coin purse for her daughter. The total value of the coins in the purse is $0.95. The number
forsale [732]
I’m guessing that this is for systems of equations?
n - nickels
d - dimes

These are the equations you start off with.
n = 5d-2
0.95 = 0.10d+0.05n

Substitute the top equation for n into the variable n in the bottom equation.
0.95 = 0.10d+0.05(5d-2)

Solve for d.
0.95 = 0.10d+0.25d-0.10
1.05 = 0.35d
3 = d

Substitute d into the top equation and solve for n.
n = 5(3)-2
n = 15-3
n = 12

There are 3 dimes and 12 nickels in the coin purse! Hope this helped <3

7 0
3 years ago
Read 2 more answers
Yana plans to run 1 mile she has run 7/10 of a mile what fraction of a mile does she have left to run
Andre45 [30]

Answer:

3/10

Step-by-step explanation:

7+3= 10 so 7+3=10/10 which is one mile

8 0
3 years ago
an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a
viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is;  \mathbf{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft  is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the  rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at  a time t and r = radius of the cone shaped at a time t

Then the similar triangles  ΔOCD and ΔOAB is as follows:

\dfrac{h}{r}= \dfrac{20}{8}    ( similar triangle property)

\dfrac{h}{r}= \dfrac{5}{2}

\dfrac{h}{r}= 2.5

h = 2.5r

r = \dfrac{h}{2.5}

The volume of the water in the tank is represented by the equation:

V = \dfrac{1}{3} \pi r^2 h

V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h

V = \dfrac{1}{18.75} \pi \ h^3

The rate of change of the water depth  is :

\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

Since the water is drained  through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

\dfrac{dv}{dt}= - 4  \ ft^3/sec

Therefore,

-4 = \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

the rate of change of the water at depth h = 10 ft is:

-4 = \dfrac{ 100 \ \pi }{6.25}\  \dfrac{dh}{dt}

100 \pi \dfrac{dh}{dt}  = -4 \times 6.25

100  \pi \dfrac{dh}{dt}  = -25

\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi}

Thus, the rate of change of the water depth when the water depth is 10 ft is;  \mathtt{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

4 0
3 years ago
Lucas is solving this problem.
kumpel [21]
I think it's d... ok?
3 0
3 years ago
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