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pentagon [3]
3 years ago
12

What is the volume of the sphere? Round your answer to two decimal places.

Mathematics
1 answer:
nekit [7.7K]3 years ago
5 0

Answer:

V = 4/3 π r 3

Step-by-step explanation:

Hope this helps!!

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The temperature was 78°F at 3 P.M. Each hour for
polet [3.4K]

Answer:

  66 °F

Step-by-step explanation:

At 4 P.M., the temperature had decreased by 3° from the temperature at 3 P.M., so was ...

  4 P.M. temp = 78 °F -3° = 75 °F

This was repeated until 7 P.M.

  5 P.M. temp = 75 °F -3° = 72 °F

  6 P.M. temp = 72 °F -3° = 69 °F

  7 P.M. temp = 69 °F -3° = 66 °F

The temperature at 7 P.M. was 66 °F.

__

Of course the repeated subtraction can be simplified using multiplication. The subtraction of 3° four times is equivalent to the subtraction of . . .

  (3°)(4) = 12°

That is, the temperature at 7 P.M. is ...

  78 °F -3°×4 = 78 °F -12° = 66 °F

6 0
2 years ago
I nee<br> d help on my fractions!!
adell [148]

Answer:

What do you need help with?

Step-by-step explanation:

8 0
3 years ago
The cost of attending an amusement park is $10 for children and $20 for adults. On a particular day, the attendance at the amuse
Lapatulllka [165]

Using a system of equations, it is found that 10,000 children attended the park that day.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

  • Variable c: Number of children in the park.
  • Variable a: Number of adults in the park.

The attendance at the amusement park is 30,000 attendees, hence:

c + a = 30,000, which is the first equation in matrix form.

Then:

a = 30,000 - c

The cost of attending an amusement park is $10 for children and $20 for adults. The total money earned by the park is $500,000, hence:

10c + 20a = 500,000, which is the second equation in matrix form.

Since a = 30,000 - c, we replace:

10c + 20a = 500,000

10c + 20(30000 - c) = 500,000

10c = 100,000

c = 10,000.

More can be learned about a system of equations at brainly.com/question/24342899

#SPJ1

8 0
2 years ago
Please help tysm ill mark brainliest
liraira [26]

Answer:

use pemdas(order of the operations)and you'll get the answer 27!!

Step-by-step explanation:

5 0
2 years ago
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

3 0
3 years ago
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