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Olin [163]
3 years ago
14

It is not always helpful to rewrite an equation in standard form as a first step in solving it ? True or false

Mathematics
2 answers:
artcher [175]3 years ago
8 0

Answer:

The answer is True.

Step-by-step explanation:


Alex3 years ago
7 0
I think it depends on the person and how you learn

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ABC is a right triangle. If AC = 7 and BC 8, find AB. Leave your answer in simplest radical form.
scoundrel [369]

Step-by-step explanation:

We don't know if AB is a hypotenuse or leg.

<u>If AB is hypotenuse:</u>

  • AB=\sqrt{AC^2+BC^2} =\sqrt{7^2+8^2} =\sqrt{49+64} =\sqrt{113}

<u>If AB is leg:</u>

  • AB=\sqrt{BC^2-AC^2} =\sqrt{8^2-7^2} =\sqrt{64-49} =\sqrt{15}
4 0
3 years ago
Read 2 more answers
Write an equation that would be perpendicular to the following equation- y= -1/5 x + 8​
brilliants [131]

Answer:

y=5x+8

Step-by-step explanation:

If its perpendicular it will have opposite slope but the same y intercept

8 0
3 years ago
Please solve fast. I will give Brainleist or whatever it’s called
alina1380 [7]

Answer:

84in

Step-by-step explanation:

84in

4 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
TICKETS:
blsea [12.9K]
You would need too do 5*5=25 if you did 5*7=35 that would be wrong
7 0
3 years ago
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