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ASHA 777 [7]
3 years ago
12

What is the sum of the arithmetic sequence 151, 137, 123,..if there are 26 terms?

Mathematics
1 answer:
ElenaW [278]3 years ago
5 0
First term: a1 = 151
common difference: d = -14 (we decrease by 14 each time, eg, 151-14 = 137)

nth term of this arithmetic sequence is...
an = a1+d(n-1)
an = 151+(-14)(n-1)
an = 151-14n+14
an = -14n+165

This will be used in the formula below

Sn = n*(a1+an)/2
<span>Sn = n*(151+(-14n+165))/2
</span><span>S26 = 26*(151+(-14*26+165))/2 ... replace every n with 26
</span>S26 = -624

The final answer here is choice C) -624
<span>
</span>
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Which is greater 9.018 or 9.108
professor190 [17]

Answer:

9.108

You only need to look at the 10nths place to determine which number is larger

Step-by-step explanation:

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3 years ago
In the diagram below, DE is parallel to XY. What is the value of y?​
nasty-shy [4]

Answer:

94

Step-by-step explanation:

Y is the alternate interior angles of the degree that measures 94. This means that they are congruent so y=94

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3 years ago
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Can someone answer this and give an explanation
Alik [6]
Answer: 27

Explanation: Bunny&Cat=10kg it can be 1/9,2/8 etc. it’s 3/7 Bunny 3 Cat 7. With that we know the Dog weighs 17 kg. 17 Dog + 7 Cat= 24. With that it’s easy, 3+7=10+17=27
6 0
2 years ago
A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min
Dmitry [639]

At any time t (min), the volume of solution in the tank is

500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}

If A(t) is the amount of salt in the tank at any time t, then the solution has a concentration of \dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}.

The net rate of change of the amount of salt in the solution, A'(t), is the difference between the amount flowing in and the amount getting pumped out:

A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)

Dropping the units and simplifying, we get the linear ODE

A'=10+5\sin\dfrac t4-\dfrac A{10}

10A'+A=100+50\sin\dfrac t4

Multiplying both sides by e^{10t} allows us to identify the left side as a derivative of a product:

10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}

\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}

e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt

Integrate and divide both sides by e^{10t} to get

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}

The tanks starts off with 30 lb of salt, so A(0)=30 and we can solve for C to get a particular solution of

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}

6 0
3 years ago
20.
Alik [6]

Answer:

After 3 seconds

Step-by-step explanation:

Given

h(t) = -16t^2 + 96t

Required

Seconds to attain maximum height

The maximum of a quadratic function y = ax^2 + bx + c is calculated using:

x = -\frac{b}{2a}

So, we have:

t = -\frac{b}{2a}

Where:

a = -16 and b = 96

So:

t = -\frac{96}{2 * -16}

t = -\frac{96}{-32}

Cancel out negatives

t = \frac{96}{32}

t = 3

5 0
2 years ago
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