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agasfer [191]
3 years ago
5

Which statement is true about the equation fraction 4 over 5z = fraction 1 over 5z − 4 + 6?

Mathematics
1 answer:
Westkost [7]3 years ago
7 0
Assuming your question asks for what values of z satisfy
\frac{4}{5z} = \frac{1}{5z} -4+6, we start by simplifying the terms  -4+6. This will get us \frac{4}{5z} = \frac{1}{5z} +2
Next, we can multiply both sides by 5z, to get 4=1+10z Subtract 1 on both sides to get 3=10z Finally, we isolate z by dividing both sides by 10. 
z=\frac{3}{10}

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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
Plz answer i will give brainliest!!!!
vichka [17]
It most likely has a 0.3 chance
6 0
2 years ago
Read 2 more answers
-8 over -2 by the power of three Minus 3 times -2 times 4
Crazy boy [7]

Answer:88


Step-by-step explanation:

You can simplify -8/-2 its 4 then take 4 to the 3rd power and you'll get 64 the take 3×-2 and you'll get -6 then multiply -6×4 and you'll get -24 then subtract 64 - (-24) which would end up being plus a positive and you'll get 88

4 0
3 years ago
Jen picked 3/4 of a gallon of strawberries in half an hour. If she keeps picking strawberries at the same rate, how many gallons
Ratling [72]
Hello!

3/4 = 0.75 as a decimal

We know that Jen picks 0.75 of a gallon of strawberries in half an hour, and now we want to find out how many gallons she'll pick in 2 hours.

Half an hour = 0.5 2 ÷ 0.5 = 4 There are four half hours in 2 hours.

0.75 × 4 = 3

ANSWER:

Jen will have picked 3 gallons of strawberries in 2 hours.
6 0
3 years ago
What is the solution to the equation 2 (x-3) ^2 =13 help help ASAP !
olganol [36]

Answer:

3+sqrt(13/2)

3-sqrt(13/2)

Step-by-step explanation:

2 (x-3) ^2 =13

Divide each side by 2

2/2 (x-3) ^2 =13/2

(x-3) ^2 =13/2

Take the square root of each side

sqrt((x-3) ^2) =±sqrt(13/2)

x-3 = ±sqrt(13/2)

Add 3 to each side

x-3+3 = 3±sqrt(13/2)

x = 3±sqrt(13/2)

4 0
3 years ago
Read 2 more answers
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