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stepan [7]
3 years ago
10

What is the first operation you start with when solving

Mathematics
1 answer:
Alex17521 [72]3 years ago
5 0
The correct answer is B addition
You will add 4 to both sides then the equation will be 2x=5
You might be interested in
What does the number line above represent?
sattari [20]
It is the last one. A closed circle means less than or equal to, so that crosses out the top two. The third says five is greater and we know that is not true because the line is decreasing in value. It must be the fourth.
8 0
3 years ago
What is the solution of the inequality shown<br> below?<br> x-1 &lt; -10
Serhud [2]

Answer:

x<-9

Step-by-step explanation:

x−1<−10

Add 1 to both sides.

x<−10+1

Add −10 and 1 to get −9.

x<−9

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Hoped this helped :)

8 0
2 years ago
Read 2 more answers
Question 4<br> Choose all numbers that make the inequality true.<br> -1 &gt;p
devlian [24]

Answer:

0 and above

Step-by-step explanation:

it can go onto infinity put 0 or Amy positive number and you got it

3 0
3 years ago
200 centimeters to 5 meters write ratio in simplest form
ss7ja [257]

Answer:

5.2

have a nice day

5 0
3 years ago
Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
Shkiper50 [21]

Answer:

\displaystyle  -\frac{1}{2} \leq x < 1

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0

Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

(6x^3+14x^2+10x+12)

is always positive and doesn't affect the result. It can be neglected. The expression

(x-\frac{1}{2})^2

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0

For the expression to be negative, both signs must be opposite, that is

(x+\frac{1}{2})\geq 0, (x-1)

Or

(x+\frac{1}{2})\leq 0, (x-1)>0

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

\displaystyle  -\frac{1}{2} \leq x < 1

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

\boxed{\displaystyle  -\frac{1}{2} \leq x < 1 }

7 0
2 years ago
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