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2 years ago

Tell whether the sequence is arithmetic. If the sequence is arithmetic, write a function rule to represent it.

Mathematics

1 answer:

fredd [130]2 years ago

8 0

Answer:

A. no

Step-by-step explanation:

I hope it help............

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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

What is 6*6? <br> and what does"*" mean?<br> the star is a asterisk but what is it for i don't know

oee [108]

Answer:

* means to multiply so 6*6 is 36

5 0

4 years ago

A plumbers plastic pipe is 4 m long, has an inside diameter of 4.0 cm and an outside diameter of 5.0cm. What is the volume of th

SCORPION-xisa [38]

Answer: $V=0.0028278\ m^3$

Step-by-step explanation:

Given

Length of the pipe $l=4\ m$

Inside diameter of the pipe $d_i=4\ cm$

Outside diameter of the pipe $d_o=5\ cm$

Volume of the pipe

$\Rightarrow V=\dfrac{\pi }{4}[d_o^2-d_i^2]\\\\\text{Insert the values}\\\\\Rightarrow V=\dfrac{\pi}{4}[5^2-4^2]\times 10^{-4}\times 4\\\\\Rightarrow V=28.278\times 10^{-4}\ m^3\\\\\Rightarrow V=0.0028278\ m^3$

5 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$