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lana66690 [7]
2 years ago
9

PLS HELP GIVEING BRANLIEST

Mathematics
2 answers:
Umnica [9.8K]2 years ago
8 0

Answer:

25

Step-by-step explanation:

2 to the power of 4 + 16

3 to the power of 2 = 9

so, 16 + 9 is 25...

7nadin3 [17]2 years ago
3 0

Step-by-step explanation:

{2}^{4 + }  {3}^{2}

=2x2x2x2+3x3

=16+9

=25

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What is the relation between the trigonometric ratios tan(R) and tan(J)? Please answer ASAP!
Ilya [14]

Step-by-step explanation:

∆PQR similar to ∆ JKL

JK/KL= PQ/QR

6/12=3/QR

QR=3×2= 6

therefore, tan(R) = PQ/QR = 3/6=1/2

and, tan(J) = KL/JK =12/6=2

\frac{ \tan(R) }{ \tan(J) }  =  \frac{ \frac{1}{2} }{2}

\frac{ \tan(R) }{ \tan(J) }   =  \frac{1}{4}

tan(J)=4.tan(R)

option D

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3 years ago
What is the answer?please ​
alexdok [17]
The answer is a^2-64
6 0
3 years ago
F(x)=x^3+3
lawyer [7]

Answer:

  (a)   -3/4

  (b)  -0.75

  (c)  -0.75

Step-by-step explanation:

It's a bit hard to tell what constitutes an "iteration" when using the bisection method to approximate a polynomial root. For the purpose here, we'll say one iteration consists of ...

  • evaluating the function at the midpoint of the bracketing interval
  • choosing a smaller bracketing interval
  • identifying the x-value known to be closest to the solution

Thus, the result of the iteration consists of a bracketing interval and the choice of one of the interval's ends as the solution approximation.

__

(a) We observe that the graphs intersect in the interval (-1, 0). For the first iteration, we evaluate f(x)-g(x) at x=-1/2. This tells us the solution is in the interval (-1, -1/2). The x-value closest to the root is x=-1/2.

For the second iteration, we evaluate the function f(x)-g(x) at x=-3/4. This tells us the solution is in the interval (-1, -3/4). The x-value closest to the root is x=-3/4.

For the third iteration, we evaluate the function f(x)-g(x) at x=-7/8. This tells us the solution is in the interval (-7/8, -3/4). The x-value closest to the root is x=-3/4.

__

(b) The graph tells us the solution is approximately 0.7549. Rounded to 2 decimal places, the solution is approximately 0.75.

__

(c) The above solution found after 3 iterations rounded to 2 decimal places is exactly 0.75.

__

See the attached table for function values.

_____

<em>Comment on bisection iteration</em>

Since you cut the interval containing the root in half with each iteration, you gain approximately one decimal place for each 3 iterations. When the function value is very nearly zero at one of the interval endpoints, it can take many more iterations to achieve a better result.

Here, it takes 4 more iterations before an x-value becomes closer to the solution (x≈-97/128). And it takes one more iteration to move the end of the interval away from -3/4. After these 5 more iterations (8 total), the solution is known to lie in the interval (-97/128, -193/256). The corresponding solution approximation is -193/256. It is still only correct to 2 decimal places.

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Symmetric property is the answer ;)
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A circle with the radius of 6
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