All categories

3 years ago

I need help with this question... 2x + 3 ≤ 13

Mathematics

2 answers:

jenyasd209 [6]3 years ago

3 0

Answer: x ≤ 5

Step-by-step explanation:

amid [387]3 years ago

3 0

X is greater than or equal to 5. you can subtract 3 from each side of the inequality, and then divide by 2. you then have the answer.

You might be interested in

Right triangle abc has one leg of length 6 cm, one leg of length 8 cm and a right angle at

Sauron [17]

Answer:
side length of the square = 10 cm

Explanation:
The attached image shows a diagram representing the scenario described in the problem.
Taking a look at this diagram, we can note that the side length of the square is equal to the hypotenuse of the right-angled triangle
Therefore, we can get the side of the square by calculating the length of the hypotenuse using Pythagorean theorem as follows:
(hypotenuse)² = (length of first leg)² + (length of second leg)²
(hypotenuse)² = (8)² + (6)²
(hypotenuse)² = 64 + 36
(hypotenuse)² = 100
hypotenuse = √100
hypotenuse = 10 cm

This means that the length of the side of the square is also 10 cm

Hope this helps :)

6 0

3 years ago

Please Show Your Work All My Points Please Help Me!!!!! 6x^2+8x-6=2-2x-2

Juliette [100K]

Answer:

6x^2+10x-6=0

Step-by-step explanation:

First, you need to add all like terms:

6x^2+8x-6=-2x

Now move the -2x to the other side:

6x^2+10x-6=0

7 0

3 years ago

Read 2 more answers

I am having trouble showing proofs question 1 is the most troubling, solving question 2 would also be most appreciated.

Alona [7]

I dont know ask mods

8 0

3 years ago

How do you solve the inequality, and graph the solution of 2n < 2

blsea [12.9K]

2n<2
divide both sides by 2 to solve/isolate n

n<1

6 0

4 years ago

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

4 years ago