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Natasha2012 [34]
3 years ago
6

Somebody plz help need help !!

Mathematics
1 answer:
kolezko [41]3 years ago
8 0
The answer would be  (3,2). hope that helped
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Please help me :((((
Phoenix [80]

Answer:

Dannielle's mistake was that she added the exponents,

x^3 + 5x^3, doesn't equal 6x^6

Step-by-step explanation:

the correct work is below:

x^3 + 5x^3 + 5x^2 - 3x^2 + 2x - 7 - 1

6x^3 + 2x^2 + 2x -8

4 0
3 years ago
Change both fractions to improper fractions. Then, do KCF.<br><br> I NEED HELP
erik [133]

Answer:

81/42

Step-by-step explanation:

8 1/10 ÷ 4 1/5

81/10 ÷ 21/5

81/10 * 5/21

(81*5)/(10*21)

405/210

81/42

8 0
3 years ago
Determine the range for the function y = (x - 3 )^2 + 4.
Aleksandr [31]

Answer:

<h3>Solutions are x=2 and x=−2 . At these points the function has vertical asymptotes. To address the range, let's first transform...</h3>

7 0
3 years ago
3 3/4+7 7/8 please help
Artemon [7]

Answer:

11.625

Hope this helps you!

3 0
3 years ago
Read 2 more answers
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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