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3 years ago

PLZ HELP MEEEEEEEEEEEEEEEEEE

Mathematics

2 answers:

adoni [48]3 years ago

6 0

Answer:

Step-by-step explanation:

its makes most sense

Oksanka [162]3 years ago

4 0

Answer:

c. because i look it ☝ done plzzz help me

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Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,

In-s [12.5K]

Answer:

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

Step-by-step explanation:

Given that:

$\iiint_W (x^2+y^2) \ dx \ dy \ dz$

where;

the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0$

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

7 0

3 years ago

If J is between H and K and HJ = 3(x + 2), JK = 3x -4 and KH = 44. What is the value of HJ?

Ira Lisetskai [31]

KH = 3(x + 2) + <span>3x - 4
KH = 3x + 6 + 3x - 4
KH = 6x + 2 = 44
6x = 42
x = 7
HJ = 3(x + 2)
HJ = 3(7 + 2)
HJ = 27</span>

7 0

3 years ago

How can logarithms be graphed with different bases?

yan [13]

You could use change of base formula

7 0

3 years ago

(-4,-6) and (3,-7) find the distance between each pair of points

Vlad [161]

Answer:

$5\sqrt{2}$

Step-by-step explanation:

distance between two points:

$d = \sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2}}$

we have:

(-4, -6), (3, -7)

$x_{1} = -4$

$y_{1} =-6$

$x_{2} = 3$

$y_{2} =-7$

so we have:

$d =\sqrt{(3-(-4))^{2}+ (-7-(-6))^{2}}\\\\d =\sqrt{(7)^{2}+ (-1)^{2}}\\\\d = \sqrt{49+ 1}\\\\d=\sqrt{50}\\\\d=5\sqrt{2}$

4 0

3 years ago

Read 2 more answers

6th grade math :) only 3 questions

Studentka2010 [4]

Answer:

1.30%

2. 30 questions

3. 60%

Step-by-step explanation:

4 0

3 years ago