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2 years ago

A tree is 78 feet tall and grows 7 feet per year. A second tree is 111 feet tall and grows 4 feet per year.

2 answers:

7 0

X = number of years
78 + 7x = 111 + 4x
-78 from both sides
7x = 33 + 4x
Subtract 4x from both sides
3x = 33
Divid both sides by 3
X= 11
In 11 years both trees will be the same height.

grigory [225]2 years ago

3 0

In 11 year's each tree will grow to the same height

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A function is shown below.<br> g(x) = 19.6 + 1.74x. what is the value of g (20)?

konstantin123 [22]

Answer:

54.4

Step-by-step explanation:

g(20) = 19.6 + 1.74 (20)

g(20) = 19.6 + 34.8

g(20) = 54.4

3 0

2 years ago

A 41 gram sample of a substance that's used to detect explosives has a k-value of 0.1392. Find the substance's half life, in day

-BARSIC- [3]

Answer:

The substance half-life is of 4.98 days.

Step-by-step explanation:

Equation for an amount of a decaying substance:

The equation for the amount of a substance that decay exponentially has the following format:

$A(t) = A(0)e^{-kt}$

In which k is the decay rate, as a decimal.

k-value of 0.1392.

This means that:

$A(t) = A(0)e^{-0.1392t}$

Find the substance's half life, in days.

This is t for which $A(t) = 0.5A(0)$ . So

$A(t) = A(0)e^{-0.1392t}$

$0.5A(0) = A(0)e^{-0.1392t}$

$e^{-0.1392t} = 0.5$

$\ln{e^{-0.1392t}} = \ln{0.5}$

$-0.1392t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.1392}$

$t = 4.98$

The substance half-life is of 4.98 days.

5 0

2 years ago

A 1 newton force will stretch a spring 1 meter. The spring/mass system is damped by a force that is 8 times the instantaneous ve

sukhopar [10]

Answer:

Step-by-step explanation:

Given the mass is m =16kg, and 1N force will stretch the spring 1 m.

That is, F =1N,Z =1m. Now find the spring constant k:

F = kL = 1 = k(1) = k= 1N/m.

The damping force is 8times the instantaneous velocity, this means β = 8,

and the external force is f(t) = 0

Initially the object compressed 0.6m above equilibrium position,

with the downward velocity is 2m/s.

The differential equation for a spring mass system with

damping force and extemal force is: mx" + βxt + kx = f(t).

so, 16x"+ 8x' + x= 0, x(0} = -0.6, x'(0)= 2m/s.

Now solve the DE:

The auxilary equation for the homogeneous equation is 16x"+8x'+x=0

solving we get, 16r² + 8r + 1 = 0 => (4r + 1)² = 0 => r = - 1/4.

Then the general solution for the homogenous system is: $x(t)=c_1e^{-t/4} +c_2te^{-t\4}$ .

Use the initial conditions x (0) = -0.6, x'(0) = 2m/s:

$x(0)=c_1e^{0} +c_2(0)e^{0}=-0.6=c_1\\x'(t)=-\frac{1}{4}c_1e^{-t/4}+c_2e^{-t/4}-\frac{1}{4}c_2te^{-t/4}\\x'(0)=-\frac{1}{4}c_1e^0+c_2e^0-\frac{1}{4}c_2(0)e^0=2=-\frac{1}{4}(-0.6)+c_2=c_2=1.85$ .