Please solve the inequality and give the steps needed to solve: x²-13x+36 ≤ 0

Mathematics

1 answer:

Svetlanka [38]2 years ago

6 0

I hope y’all have a good day!!

You might be interested in

At the beginning of an experiment, the temperature of a solution was 48°C. During a 6-hour

ExtremeBDS [4]

9514 1404 393

Answer:

-2.2(6) +1.8(4) . . . temperature change
42 °C

Step-by-step explanation:

At the end of the first 6 hours, the temperature change was ...

(-2.2 K/h)(6 h) = -13.2 K

The change in the next 4 hours was ...

(1.8 K/h)(4 h) = 7.2 K

Then the total change over the 10-hour period was ...

(-2.2)(6) +(1.8)(4) . . . . . expression for temperature change

This evaluates to -13.2 +7.2 = -6.

The final temperature is the sum of the initial temperature and the temperature change:

final temperature = 48 °C -6 °C = 42 °C

_____

Additional comment

When temperature is measured on the Celsius scale, a change of 1 degree is called a kelvin (K).

8 0

3 years ago

HELLPPPP!! GIVING ALL MY POINTS FOR THIS QUESTION! Identify m∠AMB.

Arlecino [84]

Answer:

I think it is the 3 one

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

M is the midpoint of AN, A has coordinates

maksim [4K]

Step-by-step explanation:

According to Question , M is the midpoint of AN, A has coordinates (-7,4), and M has coordinates (-4, -1).

Figure :-

$\setlength{\unitlength}{1 cm}\begin{picture}(20,12) \put(4,0.2){\line(0,-1){0.4}}\put(1,0){\line(1,0){6}} \put(3.8,-0.6){$\bf M(-4,-1) $} \put(1,-0.6){$\bf A (-7,4)$} \put(6.8,-0.6){$\bf N(x,y)$} \end{picture}$

Let the coordinates of N be ( x , y )

Now , according to Midpoint Formula , the midpoint of points say A(x , y) and B (x' , y') is given by ,

$\boxed{\blue{ \bf Midpoint =\bigg( \dfrac{x+x'}{2} , \dfrac{y+y'}{2} \bigg) }}$

On using this formula ,

$=> Midpoint = \bigg( \dfrac{x+x'}{2} , \dfrac{y+y'}{2} \bigg) \\ \\ => (-4, 1) = \bigg( \dfrac{x -7 }{2} , \dfrac{y + 4 }{2}\bigg) \\ \\ => -4 = \dfrac{x-7}{2} \\\\ => x - 7 = -8 \\\\ => x = 7 -8 \\\\ \boxed{\red{\sf=> x = -1 }} \\\\ => \dfrac{y+4}{2}=-1 \\\\=> y +4 = -2 \\ \\ y = -4-2 \\\\\boxed{\red{\sf=> y = -6 }}$