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elena55 [62]
2 years ago
6

80= 720 divided by what

Mathematics
2 answers:
dezoksy [38]2 years ago
5 0

Answer:

9

Step-by-step explanation:

8 times 9 is 72

80 times 9 is 720

720 divided by 9 is 80

Salsk061 [2.6K]2 years ago
4 0

Answer: 9

Step-by-step explanation:

9 x 80 =720

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Help me please ......
Kruka [31]
Since the block is a cube, the dimentions from corner to corner equal all the same, as long as the pieces are equal, which in this case, they are. :) So basically what you need to do is find the perimeter of one side of the cube, and that's the answer.
8 0
2 years ago
Please help me with this. It needs to be done soon, so I’d appreciate anyone who answers this.
creativ13 [48]

Answer:

A. 4 1/2 ==+ 3

B. 2 + 4

C. 1/2 + 1/2

D. 2 1/2 + 1

E. 4 1/2 + 2

Step-by-step explanation:

Read the coordinates and just like figure out the order which it's like 1 through 5 but wit fractions. hope this helps.

8 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
One of the five cards is chosen at random. Find the probability of choosing a card with a number that
svetoff [14.1K]

Answer:

<h2>Zero.</h2>

Step-by-step explanation:

Well, assuming that the cards are numbered 1, 2, 3, 4 and 5, then the probability of getting 6 is zero, because there's no card with such number.

Remember that a simple probability is a ratio between the possible events and the total number of outcomes. In this case, the total number of outcomes is 5 and the events are zero, because there's no card with 6.

8 0
3 years ago
Find the surface area of the regular pyramid
kicyunya [14]

Pyramid Surface Area = (½ * Perimeter of Base * Slant Height) + Base Area

Pyramid Surface Area = (.5 * 56 * 11) + 14 * 14Pyramid Surface Area = 308 + 196Pyramid Surface Area =504 square millimeters

Source:http://www.1728.org/volpyrmd.htm

5 0
3 years ago
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