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Radda [10]
3 years ago
15

About 57 people rent skates at the skating rink each day. It cost $4.25 to rent skates. About how much does the Rink make in ska

te rentals in one week?
Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:1695.75

Step-by-step explanation:

We know that there are 7 days in a week. Since 57 people rent skates each day, we know that 57*7=399 skates are rented each week. Since one skate cost 4.25, we know that 399*4.25 is the cost, which is 1695.75. (If you are not allowed to use a calculator, you can do 400*4.25-4.25, to make it easier to compute)

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What ia the angle of AOC and DOE
natulia [17]

(i) Angle AOC would equal 48°, as it is vertically opposite to angle DOB and they are therefore equal.


(ii) Angle DOE would equal 42°. We can see this because angles on a straight line add up to 180°, and 90 + 48 = 138, so we can do 180 - 138 = 42°.


I hope this helps!

4 0
3 years ago
Please help me with this question.
kow [346]
They are equivalent. combine like terms on the first one and 9x-4x+25 can be simplified to 5x+25.
6 0
3 years ago
Which has a different answer? A. 7 + 2 - 4 , B. 2 + 7 - 4 , C. 7 - 4 + 2 , D. 4 + 7 - 2
Sindrei [870]

Answer:

d) 4 + 7 - 2

Step-by-step explanation:

7 + 2 - 4 = 5

2 + 7 - 4 = 5

7 - 4 + 2 = 5

4 + 7 - 2 = 9

5 0
3 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
The answer to these please!
igomit [66]
\sqrt[4]{(4a^{2})^{6}} = \sqrt[4]{4096a^{12}} = 8a^{3}

\sqrt[7]{(b^{\frac{5}{3}})^{8}} = \sqrt[7]{b^\frac{40}{3}} = b^{\frac{40}{21}} = \sqrt[21]{b^{40}} = \sqrt[21]{b^{21} * b^{19}} = \sqrt[21]{b^{21}} \sqrt[21]{b^{19}} = b\sqrt[21]{b^{19}}

\sqrt[3]{\frac{c^{9}}{c^{3}}} = \sqrt[3]{c^{6}} = c^{2}

\sqrt[3]{27d^{6} * 8d^{-4}} = \sqrt[3]{27d^{6} * \frac{8}{d^{4}}} = \sqrt[3]{216d^{2}} = 6\sqrt[3]{d^{2}} = 6d^{\frac{2}{3}}
4 0
3 years ago
Read 2 more answers
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