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3 years ago

NEED HELP FAST 100 POINTS AND BRAINLIEST

Mathematics

2 answers:

Shtirlitz [24]3 years ago

7 0

Answer:

1)The mass of water increases.

2) Freezing point of water decreases.

3) buoyancy of objects in the water increases.

Step-by-step explanation:

Salinity in the oceanic water is attributed by the water coming from other sources like rivers carrying rocks and soil containing salt. Salinity decreases the temperature of water, making it cold and more denser. The cold water settles at the bottom of the ocean. It flows across the oceanic basins as slow currents. The objects being less in density flows over the surface of water. Hence,buoyancy of objects in water increases.

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Fittoniya [83]3 years ago

3 0

Answer: The answer is 1)The mass of water increases.

2) Freezing point of water decreases.

3) buoyancy of objects in the water increases.

Hope it helps please select me brainliest!

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d1i1m1o1n [39]

Given:

The given expression to find the nth term of the sequence is $d(n)=d(n-1) \cdot (-5)$

The first term of the sequence is $d(1)=8$

We need to determine the third term of the sequence.

Second term:

The second term of the sequence can be determined by substituting n = 2 in the nth term of the sequence.

Thus, we have;

$d(2)=d(2-1) \cdot (-5)$

$d(2)=d(1) \cdot (-5)$

$d(2)=8 \cdot (-5)$

$d(2)=-40$

Thus, the second term of the sequence is -40.

Third term:

The third term of the sequence can be determined by substituting n = 3 in the nth term of the sequence.

Thus, we have;

$d(3)=d(3-1) \cdot (-5)$

$d(3)=-40 \cdot (-5)$

$d(3)=120$

Thus, the third term of the sequence is 120.

7 0

3 years ago

Read 2 more answers

PLEASE HELP!

Inessa05 [86]

What is this 3rd grade

5 0

3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$