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Tema [17]
3 years ago
9

A certain medicine is given in an amount proportion to a patient’s body weight. Suppose a patient weighing 132 pounds requires 1

43 milligrams of medicine. What is the amount of medicine required by a patient weighing 156 pounds?
Mathematics
2 answers:
Natali5045456 [20]3 years ago
7 0

Answer:

  169 mg

Step-by-step explanation:

The ratio of medicine quantity to body weight is the same for both patients, so ...

  amount/(156 lb) = (143 mg)/(132 lb)

Multiply by 156 lb:

  amount = (143 mg)(156/132) = 169 mg

The patient requires 169 mg of medicine.

Veseljchak [2.6K]3 years ago
5 0

Answer:

160 milligrams......

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6 0
3 years ago
Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6
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:

brainly.com/question/25172004

#SPJ4

3 0
2 years ago
PLEASE HELP!!!!!!!!! WILL GIVE BRAINLIEST!!!!
wel

Answer:

1. 1023

2. 8

3. The 7th Term

Step-by-step explanation:

7 0
3 years ago
Plss solve 29 for me.with the steps<br> plsssßs I am begging u
lakkis [162]
That means that the first rectangle and the first triangle ave athe same perimiter

so therefor
perimiter=sides added up
 rectangle has 4 sides bu 2 is given so double it since the other side is same legnth
rec=x+x+4
multiply 2
2x+2x+8=4x+8=perimiter
this is equal to perimiter of triangle which is
x+9+x+5+x=3x+14
therefor
4x+8=3x+14
subtract 3x from both sides
x+8=14
subtract 8 from both sides
x=6
first one x=6





second rectangle and second triangle
s+7 and 3s
2 sides
2s+14+6s=p
8s+14=p

traignel=2s+12+2s+12+2s+12=6s+36=p
they are equal so
8s+14=6s+36
subtract 6s from both sides
2s+14=36
subtract 14 from both sides
2s=22
divide both sideds by 2
s=11






the values are
pair one:
x=6

pair 2:
s=11

5 0
3 years ago
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