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3 years ago

9

a desert has both fruit and yogurt inside altogether the mass of the desert is 185g the ratio of the mass to fruit to the mass o

f yoghurt is 2:3 what is the mass of yoghurt?

2 answers:

Studentka2010 [4]3 years ago

4 0

Fruit : yoghurt : total
2 : 3 : 2 + 3
2 : 3 : 5

5 parts = 185
1 part = 185 ÷ 5 = 37
3 parts = 37 x 3 = 111g

The mass of the yoghurt is 111g

coldgirl [10]3 years ago

3 0

Answer:

The mass of yogurt is $111\ g$

Step-by-step explanation:

we have that

The ratio of the mass of fruit to the mass of yogurt is $2:3$

and

The mass of the desert is $185\ g$

Let

x------> the mass of fruit

y------> the mass of yogurt

(x+y)----> the mass of desert

we know that

$x+y=185$ ------> equation A

$\frac{x}{y}=\frac{2}{3}$

$x=\frac{2}{3}y$ -----> equation B

Substitute equation B in equation A

$(\frac{2}{3}y)+y=185$

$\frac{5}{3}y=185$

$y=185*3/5=111\ g$

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Anton [14]

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

V= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder, $x^{2}+y^{2} =1$ and the ellipsoid, $4x^{2}+ 4y^{2} + z^{2} =64$

In polar coordinates, we know that

$x^{2}+y^{2} =r^{2}$

So, the ellipsoid gives

$4{(x^{2}+ y^{2)} + z^{2} =64$

4( $r^{2}$ ) + $z^{2}$ = 64

$z^{2}$ = 64- 4( $r^{2}$ )

z=± $\sqrt{64-4r^{2} }$

So, the volume of the solid is given by:

V= $\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta$

= $2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta$

To solve the integral take, $64-4r^{2}$ = t

dt= -8rdr

rdr = $\frac{-1}{8} dt$

So, the integral $\int\ r\sqrt{64-4r^{2} } rdr$ become

= $\int\ \sqrt{t } \frac{-1}{8} dt$

= $\frac{-1}{12} t^{3/2}$

= $\frac{-1}{12} (64-4r^{2}) ^{3/2}$

so on applying the limit, the volume becomes

V= $2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta$

= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta$

V = $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Since, further the integral isn't having any term of $\theta$ .

we will end here.

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Learn more about Volume in polar coordinate here:

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