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vovangra [49]
3 years ago
15

What is the magnitude (size) of -8.5

Mathematics
1 answer:
uysha [10]3 years ago
6 0

Answer:

8.5, because [-8.5] = 8.5

Step-by-step explanation:

A pex

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A local company manufactures netbook computers. Their profit function is given by this equation: y equal minus x to the power of
Westkost [7]
Answer by BlueSky06

The equation described above can also be written as,                    y = -x² + 100x + 4000To get the number of notebooks that will give them the maximum profit, we derive the equation and equate to zero.                  dy/dx = -2x + 100 = 0The value of x from the equation is 50. Then, we substitute 50 to the original equation to get the profit.                    y = -(50^2) + 100(50) + 4000 = 6500Thus, the maximum profit that the company makes is $6,500/day. 
Read more on Brainly.com - brainly.com/question/3586459#readmore
3 0
3 years ago
Read 2 more answers
HELP ME PLEASE!!!!!!!!
ioda

Answer:

the answer is 9×9 and -9×-9

8 0
2 years ago
Read 2 more answers
Please help simplify this...
juin [17]
Answer is 36.
First use the negative exponent rule. When a number is put to the negative power it becomes a fraction
4/3^2

Now 3 to the second power is 9
4/1/9 (4 over 1/9)
Now multiply 4 x 9 (dividing by 1/9 is the same as multiplying by 9)

4 x 9= 36
5 0
3 years ago
11.1 x 0.8 x 1.16 x 0.4
makkiz [27]
Answer: 4.12032 which is also equivalent to 4:12
3 0
3 years ago
Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.
zloy xaker [14]

Answer:

y=\frac{-3\pm\sqrt{4sin2x+1}}{2}

x={\pi}{4}

Step-by-step explanation:

We are given that

y'=\frac{2cos2x}{3+2y}

y(0)=-1

\frac{dy}{dx}=\frac{2cos2x}{3+2y}

(3+2y)dy=2cos2x dx

Taking integration on both sides then we get

\int (3+2y)dy=2\int cos 2xdx

3y+y^2=sin2x+C

Using formula

\int x^n=\frac{x^{n+1}}{n+1}+C

\int cosx dx=sinx

Substitute x=0 and y=-1

-3+1=sin0+C

-2=C

sin0=0

Substitute the value of C

y^2+3y=sin2x-2

y^2+3y-sin 2x+2=0

y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}

By using quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}

Hence, the solution y=\frac{-3\pm\sqrt{4sin2x+1}}{2}

When the solution is maximum then y'=0

\frac{2cos2x}{3+2y}=0

2cos2x=0

cos2x=0

cos2x=cos\frac{\pi}{2}

cos\frac{\pi}{2}=0

2x=\frac{\pi}{2}

x=\frac{\pi}{4}

3 0
3 years ago
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