All categories

3 years ago

What are 2 ways to solve 20=-5(-3+x)

2 answers:

s2008m [1.1K]3 years ago

5 0

1) 20 = -5(-3+x)
20/-5 = -3 + x
-4 = -3 + x
-4 + 3 = x
x = -1
2)20 = -5(-3 + x)
20 = 15 - 5x
20 - 15 = - 5x
5 = -5x
5/-5 = x
x = -1

lions [1.4K]3 years ago

3 0

1) First way:

$-5(-3+x) = 20$

=> $15 - 5x = 20$

=> $-5x = 20 -15$

=> $-5x = 5$

=> $x = 5/-5$

=> $x = -1$

2) Second way:

$-5(-3+x)=20$

=> $-3+x = 20/-5$

=> $- 3 + x = -4$

=> $x = -4 + 3$

=> $x = -1$

You might be interested in

43: PLEASE HELP Write the equation 22x-2y-10 in slope-intercept form

Gnoma [55]

Answer:

y= 11x -5

Step-by-step explanation:

I believe you rearrange it to get y on one side of the equation so,

2y= 22x-10

and then divide 2 on both sides to get,

y=11x-5

7 0

3 years ago

What is 8.275 rounded to the nearest hundredth.

miskamm [114]

Answer:

8.3

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

What are the solutions of 3x^2-x+4=0

guajiro [1.7K]

3x² -x+4=0

We can use quadratic formula
a=3, b= -1, c=4
$x= \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} 
\\ \\ x= \frac{1+/- \sqrt{1-4*3*4} }{2*3} 
\\ \\ x = \frac{1+/- \sqrt{-47} }{6} 

This equation does not have real roots, it has only 2 imaginary roots.$

$x= \frac{1+ \sqrt{47}i }{6} 
\\ \\ x= \frac{1- \sqrt{47}i }{6}$

6 0

3 years ago

I need help with this, plz help

antoniya [11.8K]

Answer:

<AFB

Step-by-step explanation:

7 0

2 years ago

Solve1 over 25 = 5^x ^+ 4. <br> x = -7/2<br> x = −6 <br> x = 9/2 <br> x = 2

Basile [38]

The equation seems to be

   1
-------- = 5 ^ (x + 4)
25

In that case, this is the solution, stept by step:

1) factor 25: 25 = 5^2

      1
=> ----------- = 5^(x + 4)
       5^2

2) invert 5^2

=> 5^(-2) = 5^(x+4)

3) Given that the bases are equal, the exponents also have to be equal:

=> - 2 = x + 4

3) transpose +4:

=> - 2 - 4 = x

=> x = - 6

Answer: - 6

6 0

3 years ago