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Ratling [72]
3 years ago
12

How would i do this problem?

Mathematics
2 answers:
Darya [45]3 years ago
6 0

Answer:

answer is a=22

follow me!

CaHeK987 [17]3 years ago
5 0

Answer:

a=22 hope it helps:))

You might be interested in
5 times 6 plus 1 times 8 plus 9 times 54 times 8 times 24 times 98 divided by 200 times 800 divided by 653 times 5782470
AveGali [126]

211512945555694/653

Decimal: 323909564403.819336

3 0
3 years ago
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
You owe your mom 10$. You pay her back 7$. Write an expression that can be used to represent this situation.
mezya [45]

Answer:

10$-7$=3$

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
How can I solve this problem?
Rus_ich [418]
(f+g)(x)=0
f (x) + g (x)=0

now,
substitute their values

x^2 -4x + x -18 = 0
x^2 -3x -18 =0
x^2 - 6x + 3x - 18 =0
x (x-6) + 3 (x-6) = 0
(x-6)(x+3) = 0


we have to find it's zeroes (or roots) now:

x-6 =0
x = 6


x +3 =0
x = -3
7 0
4 years ago
Does 23^-1 (mod 1000) exist? If yes solve it.
sweet [91]

Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).

Let <em>x</em> be the inverse. Then <em>x</em> is such that

23<em>x</em> ≡ 1 (mod 1000)

Use the Euclidean algorithm to solve for <em>x</em> :

1000 = 43×23 + 11

23 = 2×11 + 1

→   1 ≡ 23 - 2×11   (mod 1000)

→   1 ≡ 23 - 2×(1000 - 43×23)   (mod 1000)

→   1 ≡ 23 - 2×1000 + 86×23   (mod 1000)

→   1 ≡ 87×23 - 2×1000 ≡ 87×23   (mod 1000)

→   23⁻¹ ≡ 87   (mod 1000)

3 0
3 years ago
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