What is the measure of с 78 ООО 88 96

How does multiplication relate to time conversion?

anygoal [31]

That would first be a conversion factor is a number used to change one set of units to another, by multiplying or dividing. when a conversion is necessasry

8 0

3 years ago

How many litres of milk can be put in six hemispherical bowl each of diameter 35cm

Lesechka [4]

Answer:

Approximately 67.348 litres can be put in six hemispherical bowl with a diameter of 35 centimetres.

Step-by-step explanation:

The volume of a hemisphere ( $V$ ), measured in cubic centimetres, is obtained from this formula:

$V = \frac{2\pi}{3}\cdot R^{3}$

Where $R$ is the radius of the hemisphere, measured in centimetres.

We know that radius is the half of the diameter ( $D$ ), measured in centimetres, then:

$R = 0.5\cdot D$

( $D = 35\,cm$ )

$R = 0.5\cdot (35\,cm)$

$R = 17.5\,cm$

Now, we get the volume of each hemispherical bowl:

$V = \frac{2\pi}{3} \cdot (17.5\,cm)^{3}$

$V \approx 11224.649\,cm^{3}$

The total volume of six hemispherical bowl is:

$V_{T} = 6\cdot V$

$V_{T}= 6\cdot (11224.649\,cm^{3})$

$V_{T} = 67347.894\,cm^{3}$

From Physics we know that 1 litre equals 1000 cubic centimetres. Then:

$V_{T} = 67.348\,L$

Approximately 67.348 litres can be put in six hemispherical bowl with a diameter of 35 centimetres.

3 0

3 years ago

If f(X+4) =3x+4 , then find the value of f(4) please explain it clearly .

Paul [167]

$\\ \sf\longmapsto f(x+4)=3x+4$

$\\ \sf\longmapsto f(x)=3x+4-4$

$\\ \sf\longmapsto f(x)=3x$

Now

$\\ \sf\longmapsto f(4)$

$\\ \sf\longmapsto 3(4)$

$\\ \sf\longmapsto 12$

7 0

2 years ago

Kristen reads 11 pages per hour. In all, how many hours of reading will Kristen have to do

soldier1979 [14.2K]

Answer:

4

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just don’t want to waste my time in case you don’t want me to :)

6 0

2 years ago

Find the exact value of sin(-15)

koban [17]

$1)\ \ \ sin(- \alpha )=-sin \alpha \\2)\ \ \ cos2 \alpha =1-2sin^2 \alpha \ \ \ \Rightarrow\ \ \ 2sin^2 \alpha =1-cos2 \alpha \\\\ \Rightarrow\ \ \ sin \alpha = \sqrt{\frac{\big{1-cos2 \alpha}}{\big{2}}} \\\\\\sin(-15^0)=-sin15^0=-\sqrt{\frac{\big{1-cos30^0}}{\big{2}}} =-\sqrt{\frac{\big{1- \frac{ \sqrt{3} }{2} }}{\big{2}}} =-\sqrt{\frac{\big{2- \sqrt{3}}}{\big{4}}} =\\\\ =- \frac{1}{2} \sqrt{2- \sqrt{3}}$

7 0

3 years ago