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3 years ago

Can you help me find the solution to this problem 2^2+2x-4=0

Mathematics

2 answers:

kkurt [141]3 years ago

5 0

Answer:

x=0

Step-by-step explanation:

Basically you’re using PEMDAS to solve this, and 0/2 is 0

Nina [5.8K]3 years ago

3 0

Answer:

2²+2x-4=o

4-4+2x=0

2x=0

x=0/2=0

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lakkis [162]

Answer:

x= - 1 1/3

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2 years ago

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NikAS [45]

Answer:

9:0

Step-by-step explanation:

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3 years ago

By showing your work how do you multiply 1728 by 405

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3 years ago

5x+4y=16 -2x-4y=-4 <br>Show your step by step process.<br>Solve the system of linear equations.

slava [35]

Answer:

x=1 and y= 5 $\frac{1}{4}$

Step-by-step explanation:

To solve this, lets take the first equation and find the x value.

Because it is ordered in standard form, we need to convert to slope-intercept form, and isolate the y.

5x+4y=16

-5x -5x

4y=-5x+16

÷4 ÷4 ÷4

y= $\frac{-5}{4}$ +4

y=-1 $\frac{1}{4}$ +4

y= 5 $\frac{1}{4}$

Now that we have the y value, we can use this to find the x value by inserting it into either equation, since they both have y values in them. Going with the first one again, lets try it:

5x+4( $\frac{-5}{4}$ +4)=16

5x-5+16=16

5x+11=16

-11 -11

5x=5

÷5 ÷5

x=1 and y= 5 $\frac{1}{4}$

7 0

3 years ago

Pleasee Help!!! v2=?

adoni [48]

<h3>Given :</h3>

$\sf l = p(v_{2} - v_{1})$

$\tt v_{2} = ?$

<h3>Solution :</h3>

$\sf \dashrightarrow l = p(v_{2} - v_{1})$

$\sf \dashrightarrow \dfrac{l}{p} = v_{2} - v_{1}$

$\sf \dashrightarrow \dfrac{l}{p} + v_{1} = v_{2}$

$\sf \dashrightarrow v_{2} = \dfrac{l}{p} + v_{1}$

$\large \underline{\boxed{\bf{v_{2} = \dfrac{l}{p} + v_{1}}}}$

Hence, value of $\sf v_{2} = \dfrac{l}{p} + v_{1}$

4 0

3 years ago