1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ira [324]
3 years ago
12

Evaluate the expression 8.2(5)^2

Mathematics
2 answers:
Luden [163]3 years ago
8 0

Answer:

205

Step-by-step explanation:

5^2 = 25

8.2 * 25 = 205

wth ..

Alecsey [184]3 years ago
8 0

Answer:

205 hope this helps

Step-by-step explanation:

5^2= 25

8.2×25= 205 hope

You might be interested in
Write the ratio as a fraction in lowest terms. Be sure to make necessary conversions:
agasfer [191]

Answer:12533

Step-by-step explanation:

4 0
2 years ago
Matthew worked a summer job and saved $900 each
Dmitry_Shevchenko [17]

Answer:

1400

Step-by-step explanation:

hope that helps <3

5 0
2 years ago
A bag contains 8 blue marbles and 6 green marbles.A blue marble is removed and not replaced.Calculate the probability that the s
Solnce55 [7]

Answer:

a. Blue

Step-by-step explanation:

The answer is blue because there are still more blue than green and the ratio is 7:6. So the probability is the second ball removed is a blue ball. Hope this helps :)

7 0
3 years ago
(1+tanx)/(sinx+cosx)=secx
jek_recluse [69]

\frac{1+\tan x}{\sin x+\cos x}=\sec x is proved

<h3><u>Solution:</u></h3>

Given that,

\frac{1+\tan x}{\sin x+\cos x}=\sec x  ------- (1)

First we will simplify the LHS and then compare it with RHS

\text { L. H.S }=\frac{1+\tan x}{\sin x+\cos x}  ------ (2)

\text {We know that } \tan x=\frac{\sin x}{\cos x}

Substitute this in eqn (2)

=\frac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}

On simplification we get,

=\frac{\frac{\sin x+\cos x}{\cos x}}{\sin x+\cos x}

=\frac{\sin x+\cos x}{\cos x} \times \frac{1}{\sin x+\cos x}

Cancelling the common terms (sinx + cosx)

=\frac{1}{c o s x}

We know secant is inverse of cosine

=\sec x=R . H . S

Thus L.H.S = R.H.S

\frac{1+\tan x}{\sin x+\cos x}=\sec x

Hence proved

5 0
3 years ago
Particle P moves along the y-axis so that its position at time t is given by y(t)=4t−23 for all times t. A second particle, part
sergey [27]

a) The limit of the position of particle Q when time approaches 2 is -\pi.

b) The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2.

c) The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}.

<h3>How to apply limits and derivatives to the study of particle motion</h3>

a) To determine the limit for t = 2, we need to apply the following two <em>algebraic</em> substitutions:

u = \pi t (1)

k = 2\pi - u (2)

Then, the limit is written as follows:

x(t) =  \lim_{t \to 2} \frac{\sin \pi t}{2-t}

x(t) =  \lim_{t \to 2} \frac{\pi\cdot \sin \pi t}{2\pi - \pi t}

x(u) =  \lim_{u \to 2\pi} \frac{\pi\cdot \sin u}{2\pi - u}

x(k) =  \lim_{k \to 0} \frac{\pi\cdot \sin (2\pi-k)}{k}

x(k) =  -\pi\cdot  \lim_{k \to 0} \frac{\sin k}{k}

x(k) = -\pi

The limit of the position of particle Q when time approaches 2 is -\pi. \blacksquare

b) The function velocity of particle Q is determined by the <em>derivative</em> formula for the division between two functions, that is:

v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}} (3)

Where:

  • f(t) - Function numerator.
  • g(t) - Function denominator.
  • f'(t) - First derivative of the function numerator.
  • g'(x) - First derivative of the function denominator.

If we know that f(t) = \sin \pi t, g(t) = 2 - t, f'(t) = \pi \cdot \cos \pi t and g'(x) = -1, then the function velocity of the particle is:

v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}

v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}

The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2. \blacksquare

c) The vector <em>rate of change</em> of the distance between particle P and particle Q (\dot r_{Q/P} (t)) is equal to the <em>vectorial</em> difference between respective vectors <em>velocity</em>:

\dot r_{Q/P}(t) = \vec v_{Q}(t) - \vec v_{P}(t) (4)

Where \vec v_{P}(t) is the vector <em>velocity</em> of particle P.

If we know that \vec v_{P}(t) = (0, 4), \vec v_{Q}(t) = \left(\frac{2\pi\cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, 0 \right) and t = \frac{1}{2}, then the vector rate of change of the distance between the two particles:

\dot r_{P/Q}(t) = \left(\frac{2\pi \cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, -4 \right)

\dot r_{Q/P}\left(\frac{1}{2} \right) = \left(\frac{2\pi\cdot \cos \frac{\pi}{2}-\frac{\pi}{2}\cdot \cos \frac{\pi}{2} +\sin \frac{\pi}{2}}{\frac{3}{2} ^{2}}, -4 \right)

\dot r_{Q/P} \left(\frac{1}{2} \right) = \left(\frac{4}{9}, -4 \right)

The magnitude of the vector <em>rate of change</em> is determined by Pythagorean theorem:

|\dot r_{Q/P}| = \sqrt{\left(\frac{4}{9} \right)^{2}+(-4)^{2}}

|\dot r_{Q/P}| = \frac{4\sqrt{82}}{9}

The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}. \blacksquare

<h3>Remark</h3>

The statement is incomplete and poorly formatted. Correct form is shown below:

<em>Particle </em>P<em> moves along the y-axis so that its position at time </em>t<em> is given by </em>y(t) = 4\cdot t - 23<em> for all times </em>t<em>. A second particle, </em>Q<em>, moves along the x-axis so that its position at time </em>t<em> is given by </em>x(t) = \frac{\sin \pi t}{2-t}<em> for all times </em>t \ne 2<em>. </em>

<em />

<em>a)</em><em> As times approaches 2, what is the limit of the position of particle </em>Q?<em> Show the work that leads to your answer. </em>

<em />

<em>b) </em><em>Show that the velocity of particle </em>Q<em> is given by </em>v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t +\sin \pi t}{(2-t)^{2}}<em>.</em>

<em />

<em>c)</em><em> Find the rate of change of the distance between particle </em>P<em> and particle </em>Q<em> at time </em>t = \frac{1}{2}<em>. Show the work that leads to your answer.</em>

To learn more on derivatives, we kindly invite to check this verified question: brainly.com/question/2788760

3 0
2 years ago
Other questions:
  • A mass attached to a spring is pulled toward the floor so that its height above the floor is 10 mm (millimeters). The mass is th
    8·1 answer
  • What is the radius of the circle with the equation
    10·1 answer
  • (My teacher doesn't give credit with out how i did the answer btw)
    12·2 answers
  • Which graph is the solution to the system y &gt; 2x – 3 and y &lt; 2x + 4?
    14·2 answers
  • What is the formula for finding the perimeter of a triangle???
    12·1 answer
  • Find M Please help!!!! Last question worth 10 points!
    5·1 answer
  • Please help i will give brainliest
    8·1 answer
  • Stuck on these questions. any help? <br><br>(last word missing is degree)​
    10·1 answer
  • Bro HELPPPPPPPPPPPPPPPPPPPPPPPPP ;-;
    6·2 answers
  • George makes two transactions today what can be the sum of 78+ (-34) mean in terms of George bank account
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!