Please help me and explain this.

Mathematics

1 answer:

Zina [86]3 years ago

8 0

You are supposed to answer where the function, h(x), is decreasing. In other words, where is “y” decreasing.

In this graph, I believe it would be from (-3,4)U(8,infinity). I hope this helps.

You might be interested in

Thurs 14 Nov BARVEMBER

weqwewe [10]

Answer:

E155

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The expression x−y equals 0.7 if x and y have certain values. Evaluate the following expressions with the same values of x and y

lisov135 [29]

Answer:

-0.7

Step-by-step explanation:

$x-y=0.7$

You can choose any digits you wish for $x$ and $y$ but the difference must be 0.7. I chose 1.0 and 0.3.

$x=1\\y=0.3\\x-y=0.7\\\\y-x=\\0.3-1=\\-0.7$

3 0

3 years ago

Plz with steps .. it's very hard can anyone plz

liubo4ka [24]

Answer:

Step-by-step explanation:

$\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a} \qquad f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a} \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} }{\dfrac{1}{2\sqrt{x}} }\\\\$

$\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}} =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\$