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vichka [17]
4 years ago
14

4x-4=2x+8? PLS HELP! Pls answer the equation and give a step by step explanation!

Mathematics
1 answer:
Rina8888 [55]4 years ago
4 0

Answer:

x = 6

Step-by-step explanation:

We want to SOLVE the equation 4x-4=2x+8, that is, find the value of x that makes the equation true.

Start by subtracting 2x from both sides (which isolates x on the left):

2x - 4 = 8

Isolate 2x by adding 4 to both sides:

2x = 12

Dividing both sides by 2, we get x = 6.

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Five bulbs of which three are defective are to be tried in two bulb points in a dark room
iogann1982 [59]

Answer:

D 7

Step-by-step explanation:

Total no. of bulb= 5

no. of defected bulb= 3

no. of not defected bulb=2

Total no. of bulb combination = 5C2

=5!/2!(5-2)!

= 5!/2!3!

= 5×4×3×2×1/2×1×3×2×1

=120/12

=10

( since a room can be lighted with one bulb also)

total no. of bulb combination when room shall not light = 3C2

3!/2!(3-2)!

= 3!/2! 1!

= 3×2×1/2×1×1

= 6/2

=3

Now,

Total no. of trial when room shall light

=10-3

=7

Hence, number of trial when the room shall be lighted is 7 which is option d

8 0
3 years ago
Somebody!!!!!!
sveta [45]
Where is your problem?? so we I can try to answer it...
4 0
4 years ago
Read 2 more answers
The volume of the cylinder is 5275 ft and the height is 23 find the diameter please explain this is for school tomorrow
stepan [7]

\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\ \cline{1-1} V=5275\\ h=23 \end{cases}\implies 5275=\pi r^2(23)\implies \cfrac{5275}{23\pi }=r^2 \\\\\\ \sqrt{\cfrac{5275}{23\pi }}=r\implies 15.144\approx r~\hspace{10em}\stackrel{diameter=2r}{d\approx 30.288}

4 0
3 years ago
a square with an area of 144 cm² is reduced by a factor of 1/3 how long are the new sides of the square?​
olga55 [171]

Answer:

4  cm

Step-by-step explanation:

The square root of 144 is 12 (12 x 12 = 144) so we know our sides of the square are 12 cm²

12 / 3 = 4

(\frac{12}{1} x \frac{1}{3})

The sides of the new square are 4 :)

Hope this helps and is correct, have a nice day! :D

3 0
3 years ago
Solve each equation (Polynomial equations)
lesya [120]

{ {e}^{x} }^{2}  =  \frac{1}{ {e}^{2} } \times  {e}^{3x}  \\ \Leftrightarrow  { {e}^{x} }^{2}  =  {e}^{ 3x}  \times  {e}^{ - 2}  \\ \Leftrightarrow  { {e}^{x} }^{2}  =  {e}^{3x - 2}  \\ \Leftrightarrow  {x}^{2}  = 3x - 2 \\ \Leftrightarrow  {x}^{2}  - 3x + 2 = 0 \\ \Leftrightarrow (x - 1)(x - 2) = 0 \\ x = 1 \: \vee \: x = 2 \\  \\ {32}^{ {x}^{2} - 2x }  =  \frac{1}{ {4}^{x} } \\  \Leftrightarrow  {( {2}^{5} )}^{ {x}^{2} - 2x }  = {2}^{ - 2x}  \\ \Leftrightarrow  {2}^{5 {x}^{2} - 10x }  =  {2}^{ - 2x}  \\ \Leftrightarrow 5 {x}^{2}  - 10x =  - 2x \\ \Leftrightarrow 5 {x}^{2}    - 8x = 0 \\ \Leftrightarrow x(5x - 8) = 0 \\ \Leftrightarrow x = 0 \: \vee \: x =  \frac{8}{5}  \\  \\  { {e}^{x} }^{3}  =  {2}^{2x} \\ \Leftrightarrow  {x}^{3}  =  ln( {2}^{2x} )  \\ \Leftrightarrow  {x}^{3}  - 2x ln(2)  = 0 \\ \Leftrightarrow x( {x}^{2}  - 2 ln(2) ) = 0 \\ \Leftrightarrow x = 0 \:\vee \:  {x}^{2}  = 2 ln(2)  \\ \Leftrightarrow x = 0 \:\vee \: x =  \sqrt{2 ln(2) }  \:\vee \: x =  -  \sqrt{2 ln(2) }

8 0
4 years ago
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