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Anna71 [15]
3 years ago
11

What is the pythagorean theorem? x?

Mathematics
1 answer:
wariber [46]3 years ago
6 0

Answer:

A^2+b^2=c^2

Step-by-step explanation:

Pretty much the two shortest sides of a right triangle squared is equal to the length of the longer side (hypotenuse) squared

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Give three solutions of the inequality 6 − 11x < 61.
monitta
1,2,3

all would give negative results
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What is the 6 term of the following geometric sequence (42,6,6/7,6/49)
Tatiana [17]
8
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4k – 5m - 32k + 5k<br> Need help please
Step2247 [10]

Answer:

−23k−5m

Step-by-step explanation:

Collect like terms.

(4k-32k+5k)-5m(4k−32k+5k)−5m

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Can someone please help pleaseeee!!!
olasank [31]
The sum is y(4y − 1) + <span>(y + 5)2 = 4y² -y +y² +25 - 10y = 5y²-11y + 25 
the answer is </span><span>5y2 − 11y + 25

</span>
5 0
3 years ago
Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte
Mashcka [7]

Answer:

f'(x) > 0 on (-\infty ,\frac{1}{e}) and f'(x)<0 on(\frac{1}{e},\infty)

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

f'(x)>0

To find its decreasing interval :

f'(x)

2) Then let's find the critical point of this function:

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0

2.2 Solving for x this equation, this will lead us to one critical point since x'  is not defined for Real set, and x''=\frac{1}{e}≈0.37 for e≈2.72

x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0
3 years ago
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