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REY [17]
2 years ago
15

The area of the sector of a circle with a radius of 8 centimeters is 125.6 square centimeters. The estimated value of pi is 3.14

. the measure of the angle of the sector is
Mathematics
1 answer:
tangare [24]2 years ago
4 0
\bf \textit{area of a sector of a circle}\\\\
A=\cfrac{\theta\pi r^2}{360}\qquad 
\begin{cases}
\theta=\textit{angle in degrees}\\
r=radius\\
----------\\
r=8\\
A=125.6
\end{cases}
\\\\\\
360A=\theta\pi r^2\implies \cfrac{360A}{\pi r^2}=\theta\implies \cfrac{360\cdot 125.6}{\pi 8^2}=\theta
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0.144 divided by (-4) = <br><br> -12.684 divided by (-5) =
Alex73 [517]

0.144 ÷ (-4) = -0.036

-12.684 ÷ (-5) = 2.5368

8 0
2 years ago
Which number line shows the solution 1/2x-2&gt;0
Ymorist [56]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

\frac{1}{2}x-2>0

Solve for x

\frac{1}{2}x>2

x>4

The solution is the interval ------> (4,∞)

All real numbers greater than 4

In a number line, the solution is the shaded area at right of x=4 (open circle, the number 4 is not included in the solution)

see the attached figure

7 0
3 years ago
You are playing a game using the spinners shown.
Otrada [13]

Answer and Step-by-step explanation:

<u>You should pick Spinner A.</u> There is a high chance of getting red since there are three pieces of the spinner that has red on it, while Spinner B only has 2 instances of red.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

7 0
2 years ago
A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and
natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: E=k\frac{q}{r^{2} }; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: p=\frac{q}{A}; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: S=4\pi r^{2}; which gives us the surface of a sphere.

The inner surface: Si=4\pi (0.06m)^{2} = 0.04 m^{2}

The outer surface: S=4\pi (0.08m)^{2}=0.08m^{2}

Now we can calculate the charges,

Inner charge: Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC

Outer charge: Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC

Then, we are able to calculate both fields:

Inner field: Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}

Outer field:  Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

3 0
3 years ago
*20 POINTS*
Paraphin [41]

Answer:

If you want to use the Rational Zeros Theorem, as instructed, you need to use synthetic division to find zeros until you get a quadratic remainder.

P: ±1, ±2, ±3, ±6  (all prime factors of constant term)

Q: ±1, ±7              (all prime factors of the leading coefficient)

P/Q: ±1, ±2, ±3, ±6, ±1/7, ±2/7, ±3/7, ±6/7  (all possible values of P/Q)

Now, start testing your values of P/Q in your polynomial:

f(x)=7x4-9x3-41x2+13x+6

You can tell f(1) and f(-1) are not zeros since they're not = 0Now try f(2) and f(-2):

f(2)=7(16)-9(8)-41(4)+13(2)+6

       112-72-164+26+6 ≠ 0

f(-2)=7(16)-9(-8)-41(4)+13(-2)+6

       112+72-164-26+6 = 0  OK!! There is a zero at x=-2

This means (x+2) is a factor of the polynomial.

Now, do synthetic division to find the polynomial that results from

(7x4-9x3-41x2+13x+6)÷(x+2):

-2⊥ 7   -9   -41    13     6

         -14   46   -10    -6          

      7  -23   5       3     0     The remainder is 0, as expected

The quotient is a polynomial of degree 3:

7x3-23x2+5x+3

Now, continue testing the P/Q values with this new polynomial.  Try f(3):

f(3)=7(27)-23(9)+5(3)+3

      189-207+15+3 = 0  OK!!  we found another zero at x=3

Now, another synthetic division:

3⊥ 7   -23    5     3

          21   -6   -3_

    7    -2    -1    0  

The quotient is a quadratic polynomial:

7x2-2x-1  This is not factorable, you need to apply the quadratic formula to find the 3rd and 4th zeros:

x= (1±2√2)÷7

The polynomial has 4 zeros at x=-2, 3, (1±2√2)÷7

Step-by-step explanation:

7 0
3 years ago
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