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3 years ago

In only three days the temperature dropped by 24 degrees.How many degrees per day did the temperature drop?

Mathematics

1 answer:

yan [13]3 years ago

6 0

Answer: 8° per day.

Step-by-step explanation:

Based on the information above, since we are informed that in three days, the temperature dropped by 24 degrees, the number of degrees per day that the temperature dropped will be go by dividing 24° by 3 days. This will be:

= 24° / 3 days

= 8° per day

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3 years ago

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A baseball team played 130 games in a season. If it won sixteen more games

egoroff_w [7]

Answer:

The team has 92 wins and 38 losses.

Step-by-step explanation:

In a season 130 games are played by a baseball team.

Let us assume that the team wins x games and losses y games.

Therefore, we can write, x + y=130....... (1)

It is given that if the team won sixteen more games twice as many games as it lost.

So, we can conclude that x =2y+16 ......(2)

Hence, solving equations (1) and (2) we get, (2y+16) + y =130

⇒ 3y = 114

⇒ y =38

And from equation (2), x = 2×38 + 16 = 92

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3 years ago

Find the missing side. Round your answer to the nearest tenth.

nikdorinn [45]

Answer:

$sin\left(90\right)/x=sin\left(25\right)/16$

x = 37.85

Step-by-step explanation:

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2 years ago

A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la

SOVA2 [1]

Answer:

$8\pi\text{ square cm}$

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

$S=2\pi rh$

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

$16 = 4r^2 + h^2$

$16 - 4r^2 = h^2$

$\implies h=\sqrt{16-4r^2}$

Thus, the surface area of the cylinder,

$S=2\pi r(\sqrt{16-4r^2})$

Differentiating with respect to r,

$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

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3 years ago

Andy and Samantha are performing an experiment in a science lab. The number of bacteria cells recorded by Andy in his experiment

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Andy will record 6x + 5 bacteria cells, while Samantha will record 2^x + 3 bacteria cells. These will be equal at a time when:
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Therefore, the number of bacteria cells will be equal at 35, at t = 5 hours.

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3 years ago

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