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Ksju [112]
3 years ago
12

Mmmmmmmmmmmmmmmmmmmmmmmm

Mathematics
2 answers:
nignag [31]3 years ago
8 0
What are we supposed to answer??? :/
Pie3 years ago
5 0

Answer:

Where is the rest of the page

Step-by-step explanation:

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Solve 3/5x=15 <br> Please
Ilya [14]

Answer:

dude

if 3 = 15

5x = 5.5=25

Step-by-step explanation:

hope it helps

5 0
3 years ago
Graph the inequality x &gt; 3.<br><br>(Choose which numberline x &gt; 3 is)
Nitella [24]

Answer:

Its the first Graph the one with the open circle at 3 and going right.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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
Use the number line below to find a fraction that is equivalent to 1/2
Yuliya22 [10]

Answer:

4/8

Step-by-step explanation:

There are an infinite number of fractions equal to 1/2, but the number line in the question has 8 spaces, and shows 1 divided into 8 pieces. Therefore, the answer to this question is likely 4/8.

(may I ask why exactly this is put as a college level problem?)

4 0
3 years ago
Read 2 more answers
Identify the diameter of the circular base created by folding the figure into a right cone. HELP ASAP PLEASE!!
Akimi4 [234]

let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.

think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.

now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area.  Well, what is the circumference of that shaded area?

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi

well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill

3 0
3 years ago
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