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3 years ago

M∠ROS = 20 deg 15' 40", m∠SOT = 10 ° 12' 30" m∠ROT =

Mathematics

1 answer:

nlexa [21]3 years ago

7 0

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A 45 kg skydiver has a speed of 25 m/s at an altitude of 950 meters above the ground. Determine the

Montano1993 [528]

The kinetic energy possessed by the skydiver is 14, 062. 5 Joules

<h3>What is kinetic energy?</h3>

Kinetic energy of an object is defined as the energy possessed by an object due to its motion.

It is also described as the work that is required to accelerate a body of mass from rest to its given velocity.

The formula for calculating kinetic energy is expressed as;

Kinetic energy = 1/2 mv²

Where;

m is the mass of the object
v is the velocity of the object

Now, substitute the values

Kinetic energy = 1/2 (45)(25)²

Find the square

Kinetic energy = 1/2 (45)(625)

Find the product

Kinetic energy = 1/2(28125)

Find the quotient

Kinetic energy = 14, 062. 5 Joules

Hence, the value is 14, 062. 5 Joules

Learn more about kinetic energy here:

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3 0

1 year ago

If John read only 10 pages of a book in an hour, how long would it take to read all the pages of the book if it has 487 pages?

melisa1 [442]

Forty nine hours and ten minutes I’m pretty sure

4 0

3 years ago

Find the area of a triangle bounded by the y-axis, the line f(x)=9−4/7x, and the line perpendicular to f(x) that passes through

Setler79 [48]

<u>ANSWER: </u>

The area of the triangle bounded by the y-axis is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

<u>SOLUTION:</u>

Given, $f(x)=9-\frac{-4}{7} x$

Consider f(x) = y. Hence we get

$f(x)=9-\frac{-4}{7} x$ --- eqn 1

$y=9-\frac{4}{7} x$

On rewriting the terms we get

4x + 7y – 63 = 0

As the triangle is bounded by two perpendicular lines, it is an right angle triangle with y-axis as hypotenuse.

Area of right angle triangle = $\frac{1}{ab}$ where a, b are lengths of sides other than hypotenuse.

So, we need find length of f(x) and its perpendicular line.

First let us find perpendicular line equation.

Slope of f(x) = $\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}$

So, slope of perpendicular line = $\frac{-1}{\text {slope of } f(x)}=\frac{7}{4}$

Perpendicular line is passing through origin(0,0).So by using point slope formula,

$y-y_{1}=m\left(x-x_{1}\right)$

Where m is the slope and $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$

$y-0=\frac{7}{4}(x-0)$

$y=\frac{7}{4} x$ --- eqn 2

4y = 7x

7x – 4y = 0

now, let us find the vertices of triangle, one of them is origin, second one is point of intersection of y-axis and f(x)

for points on y-axis x will be zero, to get y value, put x =0 int f(x)

0 + 7y – 63 = 0

7y = 63

y = 9

Hence, the point of intersection is (0, 9)

Third vertex is point of intersection of f(x) and its perpendicular line.

So, solve (1) and (2)

$\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}$

Put x value in (2)

$\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}$

So, the point of intersection is $\left(\frac{252}{65}, \frac{441}{65}\right)$

Length of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right)$ and (0,9)

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}$

Now, length of perpendicular of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)$

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}$

Now, area of right angle triangle = $\frac{1}{2} \times \frac{252}{65} \times \frac{63}{65} \sqrt{65}$

$=\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

Hence, the area of the triangle is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

8 0

3 years ago

The circumference of a circle of diameter 3,5 cm is 11 cm. What is the circumference of a circle of diameter 42 cm? Answer ... c

andreyandreev [35.5K]

Answer:

131.95 cm

Step-by-step explanation:

The Formula for the circumference of a circle is πd where d is the diameter. Therefore you simply multiply 42 with PI to get the circumference

5 0

2 years ago

You rotate a triangle 90° counterclockwise about the origin. Then you translate its image 1 unit left and 2 units down. The vert

sergiy2304 [10]

Answer:

The coordinates of the original triangle are (2,4). (4,1) and (1,1)

Step-by-step explanation:

About the origin, I have to rotate a triangle 90° counterclockwise. Then I translate its image 1 unit left and 2 units down.

The vertices of the final image are (-5,0), (-2,2), and (-2,-1).

We need to get the vertices of the original triangle.

So, start from a triangle with vertices (-5,0), (-2,2), and (-2,-1) and do the reverse to get the original triangle.

So, we will translate the image triangle by 1 unit right and 2 units up and we will get the triangle with vertices (-5 + 1, 0 + 2), (- 2 + 1, 2 + 2) and (- 2 +1, - 1 + 2) ≡ (-4,2), (-1, 4) and (-1,1).

Now, we have to rotate this intermediate triangle by 90° clockwise.

Therefore, the coordinates of the original triangle are (2,4). (4,1) and (1,1) (Answer)

{Since, for 90° rotation the coordinates of all the vertices will interchange their location (i.e. x-value to y-value and y-value to x-value) and a negative sign will be added to the y-coordinate of the interchanged coordinates}

5 0

3 years ago