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Marina CMI [18]
3 years ago
8

Walmart is selling

Mathematics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

2 notebooks are $2.48, 7 notebooks are $8.68

Step-by-step explanation:

Unit rate is 1.24

1.24x2= $2.48

1.24x7= $8.68

You might be interested in
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
What concepts do you think belong in Geometry of mathematics?
dezoksy [38]
Fractions Area Volume length h
5 0
4 years ago
Hello! How do I find the surface area of this prism? Thanks!
Studentka2010 [4]
Answer: Choice C) 124 square cm

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

Explanation:

Let's calculate the area of the trapezoid shown
b1 and b2 are the parallel bases; h is the height of the 2D trapezoid
b1 = 2
b2 = 5
h = 1.5

A = h*(b1+b2)/2
A = 1.5*(2+5)/2
A = 1.5*7/2
A = 10.5/2
A = 5.25
The area of one 2D trapezoid is 5.25 sq cm
There are two of these trapezoids that form the base faces of the trapezoidal prism. So the total base area is 2*5.25 = 10.5 sq cm
Keep this value (10.5) in mind. We'll use it later.

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

Now onto the lateral surface area (LSA)
It turns out that the formula for the LSA is
LSA = p*d
where 
p = perimeter of the trapezoid shown
d = depth or height of the 3D trapezoid (I'm not using h as it was used earlier)
This formula works for any polygonal base. It doesn't have to be a trapezoid.

In this case the perimeter is,
p = 1.7+2+2.65+5
p = 11.35

So
LSA = p*d
LSA = 11.35*10
LSA = 113.5

Add this LSA to the base area found earlier
10.5+113.5 = 124

The total surface area is 124 square cm
8 0
4 years ago
I need help with this questions
Keith_Richards [23]
QS=7x-1
BC=5x-19
I think so...

8 0
4 years ago
Plzzzz helppppppppppppp
Sindrei [870]
#2 Is Complementary because angle 1 and 2 are equal to 90 degrees

#4 Is Adjacent. That mean front he middle point/vertex both angles are on the same side and don’t over lap.

To finish the rest i have some notes.
8 0
3 years ago
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