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vekshin1
4 years ago
5

(5/2)^2 + 3^3/4 what is the value of this expression ?

Mathematics
1 answer:
Nikolay [14]4 years ago
5 0
<span><span><span>(<span>5/2</span>)</span>^2</span>+<span><span>3^3</span>/4</span></span><span>
</span><span>=<span><span>25/4</span>+<span><span>3/</span>4-3</span></span></span><span>
</span><span>=<span><span>25/4</span>+<span>27/4</span></span></span><span>
</span><span>=13</span>
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Answer

2.25

Step-by-step explanation:

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3 years ago
PLEASE HELP ILL GIVE MEDALS AND MARK BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!! NEEDS TO BE ALEGABRA 2 MEATHOD!!!!!!!!
mars1129 [50]
Hi there,
This is the original inequality equation:
\frac{x}{x+1} \ \textless \  \frac{x}{x-1}
So, we first need to find the critical points of equality, and we can do that by switching the less than sign to an equal sign.
\frac{x}{x+1} = \frac{x}{x-1}
Now, we multiply both sides by x + 1:
x= \frac{x^{2} +x}{x-1}
Then, we multiply both sides by x - 1:
x^{2} -x= x^{2} +x
Next, we subtract x² from both sides:
-x=x
After that, we solve for x. We do this by adding -x to both sides and dividing by 2. Doing so gives us x = 0, which is our first critical point. We need to find a few more critical points by testing x = -1 and x = 1. Here is how we do that:
<span>x = <span>−1 </span></span>(Makes left denominator equal to 0)<span>x = 1   </span>(Makes right denominator equal to 0)Check intervals in between critical points. (Test values in the intervals to see if they work.)<span>x <<span>−1    </span></span>(Doesn't work in original inequality)<span><span><span>−1 </span>< x </span><0  </span>(Works in original inequality)<span><span>0 < x </span>< 1 </span>(Doesn't work in original inequality)<span>x > 1  </span><span>(Works in original inequality)
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3 years ago
Based on the extreme value theorem, what is the maximum value of f(x) = –x2 + 6x over the interval [1, 4]?
Tju [1.3M]

Answer:

Maximum\ value\ =9 ,at\ x=3

Step-by-step explanation:

From the question we are told that:

Function given

f(x) = -x^2 + 6x

Co-ordinates

(x,y)=[1, 4]

Generally the second differentiation of function is mathematically given by

-2x+6

Therefore critical point

x=3

Generally the substitutions of co-ordinate into function is mathematically given by

For 1

F(1)=-(1)^2 + 6(1)\\F(1)=5

For 4

F(4)=-(4)^2 + 6(4)\\F(4)=8

For critical point 3

F(3)=-(3)^2 + 6(3)\\F(3)=9

Therefore the maximum value of f(x) = –x2 + 6x over the interval [1, 4] is given by

Maximum\ value\ =9 ,at\ x=3

3 0
3 years ago
Read 2 more answers
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Answer:

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Step-by-step explanation:

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8 0
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Find the solution of the following equation whose argument is strictly between 270^\circ270 ∘ 270, degree and 360^\circ360 ∘ 360
Natasha2012 [34]

\rightarrow z^4=-625\\\\\rightarrow z=(-625+0i)^{\frac{1}{4}}\\\\\rightarrow x+iy=(-625+0i)^{\frac{1}{4}}\\\\ x=r \cos A\\\\y=r \sin A\\\\r \cos A=-625\\\\ r \sin A=0\\\\x^2+y^2=625^{2}\\\\r^2=625^{2}\\\\|r|=625\\\\ \tan A=\frac{0}{-625}\\\\ \tan A=0\\\\ A=\pi\\\\\rightarrow z= [625(\cos (2k \pi+pi) +i \sin (2k\pi+ \pi)]^{\frac{1}{4}}\\\\k=0,1,2,3,4,....\\\\\rightarrow z=(625)^{\frac{1}{4}}[\cos \frac{(2k \pi+pi)}{4} +i \sin \frac{(2k\pi+ \pi)}{4}]

\rightarrow z_{0}=(625)^{\frac{1}{4}}[\cos \frac{pi}{4} +i \sin \frac{\pi)}{4}]\\\\\rightarrow z_{1}=(625)^{\frac{1}{4}}[\cos \frac{3\pi}{4} +i \sin \frac{3\pi}{4}]\\\\ \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]\\\\ \rightarrow z_{3}=(625)^{\frac{1}{4}}[\cos \frac{7\pi}{4} +i \sin \frac{7\pi}{4}]

Argument of Complex number

Z=x+iy , is given by

If, x>0, y>0, Angle lies in first Quadrant.

If, x<0, y>0, Angle lies in Second Quadrant.

If, x<0, y<0, Angle lies in third Quadrant.

If, x>0, y<0, Angle lies in fourth Quadrant.

We have to find those roots among four roots whose argument is between 270° and 360°.So, that root is

   \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]

5 0
3 years ago
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