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densk [106]
3 years ago
14

Sandra and her friend went to the candy store. Each of the purchased a bag of jelly beans. Sandra's bag weighed 1.25 pounds. Her

friend's bag weighed 1.05 pounds. Who bought more candy?
Mathematics
2 answers:
Aleks [24]3 years ago
8 0
Sandra did, her bag was .20 pounds heavier than her friend’s.
uysha [10]3 years ago
3 0
Sandra she has .20 more that her friend
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3 years ago
Cameron needs to order some new supplies for the restaurant where he works. The
zvonat [6]

Answer:

598=194+8x

Step-by-step explanation:

To solve you would subtract 194 from its self and 598

598=194+8x

-194  -194

404=8x

then you would divide 8 from it's self and 404

which would make it 50.5

(but you cant buy half a package of spoons so realistically it'd be 50 or 51.

hope this helped :)

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3 years ago
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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. =)

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3 years ago
Ken Has Two Red Shirts, Two Blue Shirts, A Pair Of White Pants And A Pair Of Blue Pants. The Probability That He Will Wear A Red
insens350 [35]

There is a 2 in 4 chance for a red shirt

there is a 1 in 2 chance for blue pants

There is (I think) a 1 in 6 chance for the chosen pain to be worn.

8 0
3 years ago
20 POINTS<br><br>please help and have a good weekend
Over [174]

A = 6 cm; B = 3 cm; C = 8 cm; D = 10 cm

The Surface area of the prism = 120 cm².

<h3>How to Find the Surface Area of Triangular Prism?</h3>

Surface area = Area of A + B + C + D

The side lengths are:

A = 6 cm

B = 3 cm

C = 8 cm

D = 10 cm

Surface area of the prism = 2(area of triangular face) + area of rectangle 1 + area of rectangle 2 + area of rectangle 3

Area of triangular face = 1/2(b)(h) = 1/2(8)(6) = 24 cm

Area of rectangle 1 = (length)(width) = (10)(3) = 30 cm²

Area of rectangle 2 = (length)(width) = (6)(3) = 18 cm²

Area of rectangle 3 = (length)(width) = (8)(3) = 24 cm²

Surface area of the prism = 2(24) + 30 + 18 + 24 = 120 cm².

Learn more about Surface Area of Triangular Prism on:

brainly.com/question/16147227

#SPJ1

4 0
2 years ago
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