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Pavel [41]
2 years ago
6

What does this multiple by 38 , 51 , 64 , 77 , 90

Mathematics
1 answer:
nadezda [96]2 years ago
3 0

Answer:

It's an arithmetic sequence. Instead of multiplying, you're adding 13. The answer is Tn=13n+25

Step-by-step explanation:

If you minus 38 from 51 (51-38), you get 13 and this 13 is constant throughout.

It's the same if you do (64-51).

Therefore 13 is our difference, we call it d.

Next you use the formula Tn=a+(n-1)d and substitute.

a is the first term of our sequence therefore it's a=38 and d=13.

We don't know n which is the number of terms.

When we sub its gonna be Tn=38+(n-1)(13)

And then you solve.

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Hi guys, Can anyone help me with this tripple integral? Thank you:)
OleMash [197]

I don't usually do calculus on Brainly and I'm pretty rusty but this looked interesting.

We have to turn K into the limits of integration on our integrals.

Clearly 0 is the lower limit for all three of x, y and z.

Now we have to incorporate

x+y+z ≤ 1

Let's do the outer integral over x.  It can go the full range from 0 to 1 without violating the constraint.  So the upper limit on the outer integral is 1.

Next integral is over y.  y ≤ 1-x-z.   We haven't worried about z yet; we have to conservatively consider it zero here for the full range of y.  So the upper limit on the middle integration is 1-x, the maximum possible value of y given x.

Similarly the inner integral goes from z=0 to z=1-x-y

We've transformed our integral into the more tractable

\displaystyle \int_0^1 \int_0^{1-x} \int _0^{1-x-y} (x^2-z^2)dz \; dy \; dx

For the inner integral we get to treat x like a constant.

\displaystyle \int _0^{1-x-y} (x^2-z^2)dz = (x^2z - z^3/3)\bigg|_{z=0}^{z= 1-x-y}=x^2(1-x-y) - (1-x-y)^3/3

Let's expand that as a polynomial in y for the next integration,

= y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3

The middle integration is

\displaystyle \int_0^{1-x} ( y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3)dy

= y^4/12 + (x-1)y^3/3+ (2x+1)y^2/2- (2x^3+1)y/3 \bigg|_{y=0}^{y=1-x}

= (1-x)^4/12 + (x-1)(1-x)^3/3+ (2x+1)(1-x)^2/2- (2x^3+1)(1-x)/3

Expanding, that's

=\frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1)

so our outer integral is

\displaystyle \int_0^1 \frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1) dx

That one's easy enough that we can skip some steps; we'll integrate and plug in x=1 at the same time for our answer (the x=0 part doesn't contribute).

= (5/5 + 16/4 - 36/3 + 16/2 - 1)/12

=0

That's a surprise. You might want to check it.

Answer: 0

6 0
3 years ago
I need help with this 1st problem and if you can help me this one and with the other ones I will post them thx
guapka [62]
10 people....

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1              4                  4

Within 30, 1/3 got a part in the show....

30            1                   30
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1               3                   3    

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8 0
3 years ago
purchase 7 servings of yogurt each serving of yogurt weighs 1/4 pounds write an improper fraction as a mixed number to show the
miss Akunina [59]
7 times 1/4 = 7/4 = 1 3/4                                                                                                                                                                                                              Answer = 1 3/4
8 0
3 years ago
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Do u follo back tho lol
7 0
3 years ago
If a = -35, b = 10 cm and c = -5, verify that:
Murljashka [212]

Step-by-step explanation:

(i). a+(b+c) = (a+b)+c

-35+(10-5) = (-35+10)+(-5)

-35+5 = -25-5

-30 = -30

(ii). a×(b+c) = a×b + a×c

-35 × [10+(-5)] = -35×10 + -35×-5

-35 × (10-5) = -350 + 175

-35 × 5 = -350 + 175

-175 = -175

7 0
2 years ago
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