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Furkat [3]
3 years ago
11

Missing pattern 24,21, blank 15 blank ,blank

Mathematics
2 answers:
never [62]3 years ago
8 0
To figure out this pattern, let's see if there's a constant rule that this sequence follows.
From 24 to 21, you're subtracting 3.
Let's try n-3, and fill in the blanks as we go.
24 - 3 = 21.
21 - 3 = 18.
18 - 3 = 15.
15 - 3 = 12.
12 - 3 = 9.
It seems that n-3 works as a rule, so your rule is:
N-3.
The missing blanks are:
18, 12, and 9.
I hope this helps!
Flura [38]3 years ago
5 0
The pattern would be n-3
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You mean -2x+5y=20 if it was that's how you do it
so if you're finding the x intercept the y will turn to 0 why because you're only looking for x their's no need for y
so it will be -2x+0=20
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Every / represents a fraction
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Answer:

410.305

Step-by-step explanation:

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Find COS
ehidna [41]

\frac{9}{41}

Step-by-step explanation:

The ratio cosine tell us that

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In other words,

\cos(x)  =  \frac{adj}{hyp}

The side adjacent to angle a is 9. The hypotenuse is 41.

So

\cos(a)  =  \frac{9}{41}

6 0
3 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
A 15% tax is to be applied to a $120 pair of shoes. Find the total price.
postnew [5]

Total cost of item = $138

Explanation:

The tax percentage = 15%

cost of item = $120

sales tax = cost of item × The tax percentage

sales tax = 120 × 15%

=120 × 0.15

sales tax = $18

Total cost of item = initial cost of item + sales tax

Total cost of item = 120 + 18

Total cost of item = $138

7 0
11 months ago
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