All categories

3 years ago

Given: 3x - 2 ≤ 2x + 1.

Mathematics

1 answer:

Natasha2012 [34]3 years ago

5 0

Answer:

$\boxed{\boxed{\pink{\bf \leadsto Hence \ the\ correct \ option \ is \ 2 . }}}$

Step-by-step explanation:

A inequality is given to us and we need to find the solution set . The inequality given to us is :-

$\bf \implies 3x -2 \geq 2x + 1 \\\\\bf\implies 3x - 2x \geq 2 + 1 \\\\\bf\implies\boxed{\bf x \geq 3}$

This means that x belongs to Real Numbers. Hence final answer would be :-

$\boxed{\red{\bf { x | x \in R , x \geq 3 }}$

<h3>Hence option 2 is correct .</h3>

You might be interested in

Please help thank you.

Brilliant_brown [7]

Question 1:
For that point where all the lines intersect to be an incenter, the following must be true
4x - 1 = 6x - 5
Subtract both sides by 6x
-2x -1 = -5
Add both sides by 1
-2x = -4
Divide both sides by -2
x= 2

Question 2:
Plug in x=2 into any of the two equations.

You get 7.

Have an awesome day! :)

5 0

3 years ago

Which fraction is equivalent to 1/8

Law Incorporation [45]

The equivalent fractions are 2/16 8/64

6 0

3 years ago

Read 2 more answers

Topic: The Quadratic Formula

Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

$x= \frac{-b +- \sqrt{b^2-4ac} }{2a}$

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² = x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

$x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\= -4 or 1$

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

$x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\$

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

$x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2$

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

7 0

3 years ago

What is cos q if sec q = 2? I need help please please

nirvana33 [79]

I hope this helps you

sec Q=1/cos Q

2=1/cos Q

cos Q= 1/2

Q=30+2.pi.n

Q=330+2.pi.n

n€Z

6 0

3 years ago

Really needed !!!!!! Please

Ivahew [28]

PRT means multiply p, r, and t
2000*0.08*2=320

8 0

3 years ago

Read 2 more answers