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KengaRu [80]
3 years ago
8

Find the original price of a pair of shoes if the sale price is $63 after a 30% discount

Mathematics
1 answer:
pickupchik [31]3 years ago
4 0
Original price - discount = sales price
x - 0.3(x) = 63
0.7x = 63
x = 63/0.7
x = 90 <=== original price
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Which is equivalent to 437<br> O A 7<br> B. 12<br> OC 16<br> OD. 64<br> E. 81
soldier1979 [14.2K]
D 64 if this is right please give me brainlist
6 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
Bob made 4 1/4 pounds of fruit salad. He added 10 ounces of pineapple. How many 6-ounce servings did Bob make?
rosijanka [135]

Answer:

13 or B

Step-by-step explanation:

1 pound=16 oz

4 1/4=4.25, 4.25*16=68

68+10=78

78/6=13

6 0
3 years ago
Read 2 more answers
Click on the words to go to the picture
madam [21]
31 is the answer because you would do 5 x 3 an that's 15 and if it's squared u technically add 15 a second time
6 0
3 years ago
Read 2 more answers
LCM of two numbers is 1134 and HCF is 18. if one of the numbers is 162, find the other. number.<br>​
Brilliant_brown [7]
<h3>Answer:  126</h3>

=====================================================

Work Shown:

Let x and y be the two numbers.

We're given x = 162 and the variable y is unknown.

We're also given LCM = 1134 and HCF = 18

So,

LCM = (x*y)/HCF

1134 = 162*y/18

1134 = (162/18)y

1134 = 9y

9y = 1134

y = 1134/9

y = 126

The other number is 126

---------------------

Notice that

  • 162 = 18*9
  • 126 = 18*7

showing that 18 is the highest common factor (HCF) of the numbers 162 and 126. This partially confirms the answer.

Now let,

  • A = multiples of 162
  • B = multiples of 126

So,

  • A = 162, 324, 486, 648, 810, 972, 1134, 1296, ...
  • B = 126, 252, 378, 504, 630, 756, 882, 1008, 1134, 1260, ...

We see that 1134 is in each list of multiples and the smallest such common item. So the lowest common multiple (LCM) of 162 and 126 is 1134. This helps fully confirm the answer.

7 0
2 years ago
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