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Rudiy27
3 years ago
15

What is the ratio of melvins weight to the total weight of the two boys

Mathematics
1 answer:
Flauer [41]3 years ago
8 0
Ratio of volume is a cubed: b cubed
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Martin supera en 22 años a su hijo ernesto Expresa algebraicamente de 2 maneras distintas esta Relacion
scZoUnD [109]

you got the wrong brainly i guess

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2 years ago
Does anyone know what k+2x=3 Does anyone know the answer?
alexira [117]

Answer:

You can only be able to solve two variables, k and x based on what you input as your question

k= -2x+3

x= (3-k)/2

5 0
3 years ago
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A guy-wire is attached to a pole for support. If the angle of elevation to the pole is 67° and the wire is attached to the groun
Andrews [41]
The angle = 67<span>° and the pole is a right-angled triangle. Therefore the distance required to find (say "x) is the angle opposite the right-angle. The following equation needs to be solved:

cos 67 = 137 / x

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3 0
3 years ago
3) g(x)= x3 + x<br> h(x) = x + 4<br> Find g(0)h(0)
Arte-miy333 [17]

Answer:

goh(x)=x^3+12x^2+49x+68

Step-by-step explanation:

g(x) = x^3+x

h(x)=x+4

We have to find

goh(x)

goh(x) = g(h(x))=g(x+4)

If g(x) = x^3+x

g(x+4) = (x+4)^3+(x+4)

           =x^3+12x^2+49x+68

Hence our answer is

goh(x)=x^3+12x^2+49x+68

4 0
3 years ago
According to data from a medical​ association, the rate of change in the number of hospital outpatient​ visits, in​ millions, in
GaryK [48]

Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

3 0
3 years ago
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