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3 years ago

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Find the average temperature on that part of the plane 2x + 5y +z = 9 over the square |x| <= 1, y <= 1, where the temperat

ure is given by T(x,y,z) = e^-z. The average value is :_______

1 answer:

vivado [14]3 years ago

5 0

Answer:

This question answer is attached in the attachment,

Step-by-step explanation:

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Solve for x: x5 + x4 − 7x3 − 7x2 − 144x − 144 = 0

motikmotik

$x^5 + x^4 - 7x^3 - 7x^2 -144x - 144 = 0 \\x^4(x+1)-7x^2(x+1)-144(x+1)=0\\(x^4-7x^2-144)(x+1)=0\\\\x+1=0\Rightarrow x=-1\\\\x^4-7x^2-144=0\\x^4-16x^2+9x^2-144=0\\x^2(x^2-16)+9(x^2-16)=0\\(x^2+9)(x^2-16)=0\\(x^2+9)(x-4)(x+4)=0\\\\x^2+9=0\vee x-4=0 \vee x+4=0\\x=4 \vee x=-4\\\\x\in\{-4,-1,4\}$

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3 years ago

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How do u get the area of a circle?

galina1969 [7]

Answer: πr^2

Step-by-step explanation:

π is pie which is about 3.14

but you need to find the radius and square it is multiply it by itself (Example radius = 2 so 2x2=4)

then multiply the radius squared by π(pie) which is 3.14 or use a calculator.

7 0

3 years ago

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The probability of event A is x, and the probability of event B is y. If the two events are independent, which of these conditio

zzz [600]

If two events are independent, then

$Pr(A\cap B)=Pr(A)\cdot Pr(B)$ .

Use formulas for conditional probabilities:

$Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)},\\ \\ Pr(B|A)=\dfrac{Pr(A\cap B)}{Pr(A)}$ .

For independent events these formulas will be:

$Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}=\dfrac{Pr(A)\cdot Pr(B)}{Pr(B)}=Pr(A),\\ \\ Pr(B|A)=\dfrac{Pr(A\cap B)}{Pr(A)}=\dfrac{Pr(A)\cdot Pr(B)}{Pr(A)}=Pr(B)$ .

Now in your case $Pr(A)=x,\ Pr(B)=y$ and $Pr(A|B)=x,\ Pr(B|A)=y, Pr(A\cap B)=x\cdot y$ .

This shows that the only correct choice is A.

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4 years ago

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Add, subtract, multiply, or divide. Round answers to money problems to the nearest cent. Show how you solved these four problems

Katena32 [7]

Answer:

The formatting on this is a little weird but if I'm reading it correctly:

1) 3.38

2) 7.29

3) 0.05

4) 0.42

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3 years ago

Find the condition that one root of the quadratic equation may be 1 more than the other.

Eddi Din [679]

<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>

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4 years ago