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2 years ago

Please help with 43 as soon as possible

Mathematics

1 answer:

irakobra [83]2 years ago

8 0

Answer:

Step-by-step explanation:

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The figure below shows rectangle ABCD with diagonals segment AC and segment DB:

djverab [1.8K]

Answer:

segment BC ≅ segment BC (reflexive property of congruence)

Step-by-step explanation:

Reflexive property of congruence states that an angle, line segment, or shape is always congruent to itself.

Before statement 6, you need to prove that side BC is the same for both triangles

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2 years ago

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Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

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3 years ago

What is the inverse of the function g(x) = x^3/8 + 16 ?

lesya692 [45]

Answer:

$\hookrightarrow \: { \rm{f(x) = \frac{ {x}^{3} }{8} + 16}} \\$

• let f(x) be m:

${ \rm{m = \frac{ {x}^{3} }{8} + 16}} \\$

• make x the subject of the function:

${ \rm{8m = {x}^{3} + 128}} \\ \\ { \rm{ {x}^{3} = 8m - 128 }} \\ \\ { \rm{ {x}^{3} = 8(m - 16) }} \\ \\ { \rm{x = \sqrt[3]{8} \times \sqrt[3]{(m - 16)} }} \\ \\ { \rm{x = 2 \sqrt[3]{(m - 16)} }}$

• therefore:

${ \boxed{ \rm{ {f}^{ - 1} (x) = 2 {(m - 16)}^{ \frac{1}{3} } }} }\\$

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2 years ago

Find the mean. Z-score = -1.4 Standard Deviation: 7 x = 20

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Answer:

9.6

Step-by-step explanation:

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3 years ago

The ratio of 3 to 4 can be written in all the following ways except _______. 4/3 3/4 3:4 three-fourths

nirvana33 [79]

The ratio of 3 to 4 can be written in all the following ways except 4/3.

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