I NEED HELP ASAP ITS DUE SOON

LiRa [457]

Solving these equations can prove to be tricky, but once you understand them, ridiculously easy! Remember: your goal is to reduce it to (variable=number). Make sure to use opposite operations to cancel out numbers, and don't forget what you do to one side you must do to the next. There isn't one true order, but you'll know you have a great understanding of what you're doing if you can solve it numerous different times and wind up with the same result!

Set up the equation!

$4a-6=-2+14$

Let's try subtracting 14 from both sides! We must do the opposite operation to cancel it out.

$14-14=0\\-6-14= -20$

Now we have

$4a-20=-2$

We have to start on the left side now, because we can't have an equation with one blank side. The number you start with will end up getting canceled out, as 0. How about we add 20 to both sides?

$-20+20=0\\-2+20=18$

Now we have

$4a=18$

Next, we have to apply the opposite opperation to both sides. The only opperation left is multiplication (4a). Therefore, we need to divide by 4, because we have to isolate a.

$4a/4=a\\18/4=4.5$

Now we have

$a=4.5$

Your answer is a=4.5!

Remember: Combine like terms first, it's much easier!

Please let me know if this helped or if you have questions, and if you can, I'd be so grateful to receive a rating! Two-sided equations took me a long time, so I want to help you understand!

5 0

3 years ago

Which integer CANNOT be a solution to the inequality 6x − 4 ≤ 7x − 3 ?

mezya [45]

Answer:

0

Step-by-step explanation:

4 0

3 years ago

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

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Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

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Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

3 years ago

What are prime factors of 1000?

Ede4ka [16]

Answer:

1,2,4,5,8,10,20,25,40,50,10,125,200,250,500,1000

Step-by-step explanation:

prime factorization is

1000=2*2*2*5*5*5

4 0

2 years ago

Given m|n, find the value of x.

yaroslaw [1]

X=12

Because 180-168= 12

7 0

3 years ago