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Likurg_2 [28]
2 years ago
9

I don’t understand plsss helppp

Mathematics
1 answer:
Goshia [24]2 years ago
5 0

Answer:

gyuggbgzsZdxfghuiguyftdrtffyurtyryjjb

Step-by-step explanation:

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[|(x-4)÷(x+5)|]\leq[4]
Katena32 [7]

Answer:

Solutions are;

x = -8

x = -3.2

Step-by-step explanation:

Here, we want to solve the given equation for x

|(x-4)/(x + 5)| = 4

From what we have, this is an absolute value equation and thus, we are going to have two solutions

These are;

x-4/x+ 5 = 4

x-4 = 4(x + 5)

x-4 = 4x + 20

x-4x = 20 + 4

-3x = 24

x = -24/3

x = -8

secondly;

x-4/x+5 = -4

x-4 = -4(x + 5)

x-4 = -4x - 20

x + 4x = -20 + 4

5x = -16

x = -16/5

x = -3.2

3 0
3 years ago
I need help! asap please​
nasty-shy [4]

Answer:

-1/1

Step-by-step explanation:

it's negative 1 over one because the graph is constantly going down but still to the right side and the line is touching every corner of the boxes

6 0
2 years ago
Read 2 more answers
Two figures are shown. EFJL is dialation of ABIK AB has a length of 12 and EF has a length of 3
Dimas [21]

The scale factor of dilation from AB to EF is 0.25

<h3>How to determine the scale factor of dilation?</h3>

The given parameters are:

  • EFJL is dilation of ABIK
  • AB has a length of 12
  • EF has a length of 3

From the above parameters, side lengths AB and EF are corresponding sides

This means that:

The scale factor of dilation (k) is

k = AB/EF

So, we have

k = 3/12

Evaluate

k = 0.25

Hence, the scale factor of dilation from AB to EF is 0.25

Read more about scale factors at:

brainly.com/question/15891755

#SPJ1

4 0
1 year ago
Writing g for the acceleration due to gravity, the period,T, of a pendulum of length l is given by
ivanzaharov [21]

Step-by-step explanation:

T=2\pi\sqrt{\frac{l}{g}}

T+\Delta T=2\pi\sqrt{\frac{(l+\Delta l)}{g}} =2\pi\sqrt{\frac{l}{g}}\sqrt{1+\frac{\Delta l}{l}}=2\pi\sqrt{\frac{l}{g}}(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))=T(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))

Here, the Taylor approximation for a square root was applied, and O(x) stands for all negligible terms of Taylor's sum with respect to variable x.

So, \Delta T=T\frac{1}{2}\frac{\Delta l}{l}

b. For an increase of 2%, that is:

\frac{\Delta l}{l}=0.02

\frac{\Delta T}{T}=\frac{1}{2}0.02=0.01=1\%

7 0
3 years ago
Read 2 more answers
(This is another question, not the same!)
vova2212 [387]
The set with the strongest linear association is the top left
5 0
3 years ago
Read 2 more answers
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