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Alexeev081 [22]
3 years ago
12

The point (3,9) lies in which quadrant?

Mathematics
2 answers:
AfilCa [17]3 years ago
8 0
Answer is in this link www.gethelp.com
borishaifa [10]3 years ago
6 0

Answer:

Wheres the actual graph with the quadrants?

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An icicle with a diameter of 15.5 centimeters at the top, tapers down in the shape of a cone with a length of
Helga [31]

Answer:

Step-by-step explanation:

Note: I will leave the answers as fraction and in terms of pi unless the question states rounding conditions to ensure maximum precision.

From the question, we can tell it is a inversed-cone (upside down)

Volume of Cone = \pi r^{2} \frac{h}{3}

a) Given Diameter , d = 15.5cm and Length , h = 350cm,

we first find the radius.

r = \frac{d}{2} \\=\frac{15.5}{2} \\=7.75cm

We will now find the volume of the cone.

Volume of cone  \pi (7.75)^{2} \frac{350}{3} \\= \frac{168175\pi }{24}

We know the density of ice is 0.93 grams per 1cm^{3}

1cm^{3} =0.93g\\\frac{168175\pi }{24}  cm^{3} =0.93(\frac{168175\pi }{24} )\\= 20473 g(Nearest Gram)

b) After 1 hour, we know that the new radius = 7.75cm - 0.35cm = 7.4cm

and the new length, h = 350cm - 15cm = 335cm

Now we will find the volume of this newly-shaped cone.

Volume of cone = \pi (7.4)^{2} \frac{335}{3} \\= \frac{91723\pi }{15} cm^{3}

Volume of cone being melted = New Volume - Original volume

= \frac{168175\pi }{24} -\frac{91723\pi }{15} \\= \frac{35697\pi }{40} cm^{3}

c) Lets take the bucket as a round cylinder.

Given radius of bucket, r = 12.5cm (Half of Diameter) and h , height = 30cm.

Volume of cylinder = \pi r^{2} h\\=\pi (12.5)^{2} (30)\\=\frac{9375\pi }{2} cm^{3}

To overflow the bucket, the volume of ice melted must be more than the bucket volume.

Volume of ice melted after 5 hours = 5(\frac{35697\pi }{40} )\\=\frac{35697\pi }{8} cm^{3}

See, from here of course you are unable to tell whether the bucket will overflow as all are in fractions, but fret not, we can just find the difference.

Volume of bucket - Volume of ice melted after 5 hours

= \frac{9375\pi }{2} -\frac{35697\pi }{8 } \\=\frac{1803\pi }{8}cm^{3}

from we can see the bucket can still hold more melted ice even after 5 hours therefore it will not overflow.

4 0
2 years ago
The sine ratio of an angle is the opposite side over the adjacent side.
Agata [3.3K]

Answer:

never

Step-by-step explanation:

In a right triangle, every angle has an opposite side and an adjacent side. And there's always a hypotenuse.

The sine of an angle is the ratio of the length of the opposite side divided by the length of the hypotenuse.

The tangent ratio of an angle is the opposite side over adjacent side, it is not the sine ratio of a triangle.

For example, ABC is a right angled triangle having length of the sides are a,b,c.

So

sine\ C=\frac{a}{c}

3 0
3 years ago
What is the reciprocal of –5?
Anon25 [30]
-5 is the same as -5/1

Flip the fraction:

-1/5

2/9 / 1/4

Flip the 2nd fraction and multiply:

2/9 * 4/1

Multiply the numerators and denominators:

8/9
4 0
3 years ago
Read 2 more answers
A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If
Blababa [14]

Answer: Our required probability is \dfrac{1}{2}

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = \dfrac{1}{3}

So, Probability of getting head from fair coin = \dfrac{1}{2}

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}

Hence, our required probability is \dfrac{1}{2}

No, the answer is not \dfrac{1}{3}

5 0
3 years ago
The volume of this rectangular prism is 36 cm^2. How many ½ cm cubes will fit in this rectangular prism? Show your reasoning.
nata0808 [166]

Answer:

Liquid R has a mass of of 1 kg at a temperature of 30°c kept in a refrigerator to freeze . Given the specific heat capacity is 300 J kg-¹ °c-1 and the freezing point is 4°c . Calculate the heat release by liquid R.

Step-by-step explanation:

Liquid R has a mass of of 1 kg at a temperature of 30°c kept in a refrigerator to freeze . Given the specific heat capacity is 300 J kg-¹ °c-1 and the freezing point is 4°c . Calculate the heat release by liquid R.

7 0
3 years ago
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