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Sunny_sXe [5.5K]
3 years ago
13

The period of a compound pendulum is given by T=2π√(h²+k²/gh) express k in terms of T,h and g​

Mathematics
1 answer:
makvit [3.9K]3 years ago
8 0

Answer:

qss-qvcj-kwr join guys iam bored

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1 a-2a-{3a-(4a-5)} i need help plz and on this one to<br> 13(5-t)-(5-t)-12(5-t)
KengaRu [80]

Answer:

1. -5

2. 0

Step-by-step explanation:

<em>1.</em> Distribute the Negative Sign:

=a−2a+−1(3a−(4a−5))

=a+−2a+−1(3a)+−1(−4a)+(−1)(5)

=a+−2a+−3a+4a+−5

Combine Like Terms:

=a+−2a+−3a+4a+−5

=(a+−2a+−3a+4a)+(−5)

Answer:

=−5

<em>2. </em>Distribute:

=(13)(5)+(13)(−t)+−5+t+(−12)(5)+(−12)(−t)

=65+−13t+−5+t+−60+12t

Combine Like Terms:

=65+−13t+−5+t+−60+12t

=(−13t+t+12t)+(65+−5+−60)

Answer:

=0

8 0
3 years ago
How do you work out <br> 5x71
Gwar [14]

Answer:

355

Step-by-step explanation:

8 0
3 years ago
Please help asap i need to get an A on this i will give brainliest
murzikaleks [220]
Y=44
this is your answer add -6 and 50 to get 44.
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3 0
3 years ago
Find the surface area and volume of a cube with sides that are 6 inches.
lys-0071 [83]

Answer:

Step-by-step explanation:

Surface Area S=6s²

S=6(36)

S=216 in²

Volume V=s³

V=6³

V=216 in³

8 0
3 years ago
Please guys please pllease
klio [65]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

Let's calculate its discriminant :

\qquad \sf  \dashrightarrow \: 4 {u}^{2}  + 16u + 41 = 40

\qquad \sf  \dashrightarrow \: 4 {u}^{2}  + 16u + 41 - 40 = 0

\qquad \sf  \dashrightarrow \: 4 {u}^{2}  + 16u + 1  = 0

Here, if we equate it with general equation,

  • a = 4

  • b = 16

  • c = 1

\qquad \sf  \dashrightarrow \: disciminant =  {b}^{2}  - 4ac

\qquad \sf  \dashrightarrow \: d = (16) {}^{2} - (4 \times 4 \times 1)

\qquad \sf  \dashrightarrow \: d = (16) {}^{2} - (16)

\qquad \sf  \dashrightarrow \: d = 16(16 - 1)

\qquad \sf  \dashrightarrow \: d = 16(15)

\qquad \sf  \dashrightarrow \: d = 240

Now, since discriminant is positive ; it has two real roots ~

The roots are :

\qquad \sf  \dashrightarrow \: u =  \dfrac{ - b \pm \sqrt{ d } }{2a}

\qquad \sf  \dashrightarrow \: u =  \dfrac{ - 16\pm \sqrt{ 240 } }{2 \times 4}

\qquad \sf  \dashrightarrow \: u =  \dfrac{ - 16\pm 4\sqrt{ 15 } }{8}

\qquad \sf  \dashrightarrow \: u =  \dfrac{ 4(- 4\pm \sqrt{ 15 }) }{8}

\qquad \sf  \dashrightarrow \: u =  \dfrac{ - 4\pm \sqrt{ 15 } }{2}

So, the required roots are :

\qquad \sf  \dashrightarrow \: u =  \dfrac{ - 4 -  \sqrt{ 15 } }{2}  \:  \: and \:  \:  \dfrac{ - 4 +  \sqrt{15} }{2}

4 0
2 years ago
Read 2 more answers
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