through how many radians does the minute hand of a clock rotate in 25 minutes. Round answer to one decimal place.

HELP HELP HELP I NEED HELPPP can anyone answer this the question is on the picture it would help a lot please!!!

forsale [732]

It’s -1/2 (negative 1 over 2)

4 0

3 years ago

Quadrilateral STWR is inscribed inside a circle as shown below. Write a proof showing that angles T and R are supplementary.

AlladinOne [14]

Observe the given figure.

Here, angle T and R are the inscribed angles.

And sum of inscribed angles is half of measure of intercepted arcs.

Since, the measure of intercepted arcs is 360 degrees.

Therefore, $\angle T + \angle R = \frac{1}{2} \times 360 ^{\circ}$

$\angle T + \angle R = 180 ^{\circ}$

Hence, the sum of the two angles is 180 degrees.

Therefore, these are the supplementary angles.

Therefore, angle T and angle R are supplementary angles.

Hence, proved.

8 0

3 years ago

What are the parallel sides of a trapezoid

saul85 [17]

Answer:

Parallel means that the lines will never cross. If we look at a traditional trapezoid, the top side and the bottom side are straight lines that will never cross one another. The left and right sides are slanted towards one another, to they are not parallel.

Hope this helped. : )

3 0

3 years ago

If the boy is 5’6 tall and his shadow is 4 ft and the shadow of the pole 18 ft determine the height of the pole to the nearest 1

Dominik [7]

The answer is 24.7ft

7 0

3 years ago

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop

xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

$f'(x)=\frac{1}{3x}+10 x$

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248$

$x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012$

$x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057$

$x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052$

So, root of the equation =0.205 (Approx)

Approximate relative error

$=\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41$

Approximate relative error in terms of Percentage

=0.41 × 100

= 41 %

7 0

3 years ago