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mojhsa [17]
3 years ago
11

1) m − 3 ≥ 3 A) m ≥ −6 : B) m ≥ 9 : C) m ≥ −6 : D) m ≥ 6 :

Mathematics
1 answer:
FromTheMoon [43]3 years ago
3 0

Answer:

The answer should be A and C because they are the same question

Step-by-step explanation:

I just looked at the equations and solved it from there

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PLEASE HELP! I’ll GIVE THE BRAINLIEST TO WHOEVER ANSWERS CORRECTLY!!
wariber [46]

Answer:

its 8.5

Step-by-step explanation:

8 0
3 years ago
**Spam answers will not be tolerated**
Morgarella [4.7K]

Answer:

f'(x)=-\frac{2}{x^\frac{3}{2}}

Step-by-step explanation:

So we have the function:

f(x)=\frac{4}{\sqrt x}

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

Therefore, our derivative would be:

\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}

Place the 4 in front:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})

Distribute:

=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}

The numerator will use the difference of two squares. Thus:

=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Simplify the numerator:

=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Both the numerator and denominator have a h. Cancel them:

=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Now, substitute 0 for h. So:

=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})

Simplify:

=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

=4( \frac{-1}{(x)(2\sqrt{x})})

Multiply across:

= \frac{-4}{(2x\sqrt{x})}

Reduce. Change √x to x^(1/2). So:

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

Add the exponents:

=-\frac{2}{x^\frac{3}{2}}

And we're done!

f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}

5 0
3 years ago
What is the answer to the problem i need help with?
AveGali [126]

Answer:

C

Step-by-step explanation:

Recall the equation for a circle:

(x-h)^2+(y-k)^2=r^2, where the center is (h,k) and the radius is r.

The given equation is:

(x+5)^2+(y+7)^2=21^2

Another way to write this is:

(x-(-5))^2+(y-(-7))^2=21^2

Thus, we can see that h=-5 and k=-7.

The center is at (-5, -7).

6 0
4 years ago
Two drivers are exploring the bottom of a trench in the Pacific Ocean. Dominic is at 175 feet below the surface of the ocean and
mestny [16]

\huge\boxed{\text{Dominic}}

In this problem, 0 represents the sea level. Since Dominic is 175 feet below the surface (sea level), his current location is at -175 feet.

The same concept can be applied to Karen, who is at -138 feet.

We need to find who is lower, or farther below 0. This means the answer is \boxed{\text{Dominic}}.

7 0
4 years ago
Simplify<br> −6+{14+2[60−9(1+3)]}
ziro4ka [17]
The answer is 56.

I know this because...
<span>-6 + {14 + 2 [60 − 9(1 + 3)]}
-6 + </span>{14 + 2[60 − 9(4)]}
-6 + {14 + 2[60 − 36]}
-6 + {14 + 2[24]}
-6 + {14 + 48}
-6 + {62}
-6 + 62 = 56
6 0
3 years ago
Read 2 more answers
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