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belka [17]
3 years ago
14

How do I solve inequalities with fractions on both sides?

Mathematics
1 answer:
poizon [28]3 years ago
4 0
In this situation you would add 9 to the other side of the inequality 2/5x < 9/10+9
2/5x< 9.9
:)
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A pail holds 3 1/2 gallons of water. How much is this in cups?
rewona [7]
3.5 gallons
1 gallon = 4 quarts
1 quart = 2 pints
1 pint = 2 cups

3.5 * 4 * 2 * 2 = 56 cups
5 0
1 year ago
Translate the phrase into an algebraic expression.<br> 8 less than x
Crank

Answer:

8-x is the expression for the question

7 0
3 years ago
xion orders 5 loves of bread from the website bakery the total shipping weight is 9 pounds as model of the bread what is in poun
Ray Of Light [21]
Each loaf of bread is 1.8 pounds
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3 years ago
If f(x) = x2 − 6x + 3 x − 1 , what is f(−1)? A) −5 B) −2 C) 2 D) 5
ale4655 [162]

Answer:

f(-1)=3

Step-by-step explanation:

f(x)=x^2-6x+3x-1

f(x)=x^2-3x-1

f(-1)=(-1)^2-3(-1)-1

f(-1)=1+3-1

f(-1)=4-1

f(-1)=3

8 0
3 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.

Hence, x(t) =-0.6e^{-t/4}+1.85te^{-t/4}.

5 0
3 years ago
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