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Alinara [238K]
2 years ago
8

Please help with 43 as soon as possible

Mathematics
1 answer:
irakobra [83]2 years ago
8 0

Answer:

D

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

4 0
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}}

4 0
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.<br> Z-score = -1.4<br> Standard Deviation: 7<br> x = 20
Umnica [9.8K]

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]
<span>The ratio of 3 to 4 can be written in all the following ways except 4/3. </span>
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3 years ago
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