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ehidna [41]
3 years ago
6

Olivia and her three siblings bring a sack lunch to school each day, consisting of a bagel, an apple, a cookie, and a juice box.

When Olivia's mom goes grocery shopping, she likes to purchase the same amount of each lunch item so that she can make complete lunches, with no leftover items. However, each item comes in a different sized package, as shown below:
Bagels: six in a bag
Apples: eight in a bag
Cookies: twelve in a box
Juice Boxes: nine in a box
Find the least number of packages she must purchase in order to have the same amount of each item. Then, complete each statement.

Olivia's mom should purchase how many boxes of cookies?
Mathematics
2 answers:
stiv31 [10]3 years ago
8 0

Find the Least Common Multiple of the numbers, then divide that by the number of cookies per box. That should give you the answer. I hope that helps!

alexandr402 [8]3 years ago
6 0
She should purchase 6 boxes of cookies. That would give her 72 cookies.
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Step-by-step explanation:

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2 years ago
Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value
vodomira [7]

Answer:

\rm\displaystyle \displaystyle \displaystyle θ=    {60}^{ \circ} , {300}^{ \circ}

\rm \displaystyle a =    - \frac{   \sqrt{3} }{2}    - 1, \frac{\sqrt{3}}{2}  - 1

Step-by-step explanation:

we are given two <u>coincident</u><u> points</u>

\displaystyle  P( \sin(θ)+2,  \tan(θ)-2)   \: \text{and } \\  \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)

since they are coincident points

\rm \displaystyle  P( \sin(θ)+2,  \tan(θ)-2)    = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)

By order pair we obtain:

\begin{cases}  \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) =  \sin( \theta)   + 2 \\   \\  \displaystyle 3 \sin( \theta)  - 2  \cos( \theta)  + a =  \tan( \theta)  - 2\end{cases}

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using <u>substitution</u><u> method</u>

to do so, make a the subject of the equation.therefore from the second equation we acquire:

\begin{cases}  \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )=  \sin( \theta)   + 2 \\   \\  \boxed{\displaystyle  a =  \tan( \theta)  - 2 - 3 \sin( \theta)   +  2  \cos( \theta) } \end{cases}

now substitute:

\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta)  - 2 - 3 \sin( \theta)   +  2  \cos( \theta)   \}=  \sin( \theta)   + 2

distribute:

\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)  - 6 \sin( \theta) \cos( \theta)    + 4  \cos ^{2} ( \theta)   =  \sin( \theta)   + 2

collect like terms:

\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)     + 4  \cos ^{2} ( \theta)   =  \sin( \theta)   + 2

rearrange:

\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta)  - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + =  \sin( \theta)   + 2

by <em>Pythagorean</em><em> theorem</em> we obtain:

\rm\displaystyle \displaystyle 4  - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)  =  \sin( \theta)   + 2

cancel 4 from both sides:

\rm\displaystyle \displaystyle   - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)  =  \sin( \theta)    - 2

move right hand side expression to left hand side and change its sign:

\rm\displaystyle \displaystyle   - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2  =  0

factor out sin:

\rm\displaystyle \displaystyle  \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2  =  0

factor out 2:

\rm\displaystyle \displaystyle  \sin (θ) (- 2 \cos(θ)+1)  + 2(- 2\cos( \theta) + 1 ) =  0

group:

\rm\displaystyle \displaystyle ( \sin (θ)   + 2)(- 2 \cos(θ)+1)  =  0

factor out -1:

\rm\displaystyle \displaystyle -  ( \sin (θ)   + 2)(2 \cos(θ) - 1)  =  0

divide both sides by -1:

\rm\displaystyle \displaystyle   ( \sin (θ)   + 2)(2 \cos(θ) - 1)  =  0

by <em>Zero</em><em> product</em><em> </em><em>property</em> we acquire:

\begin{cases}\rm\displaystyle \displaystyle   \sin (θ)   + 2 = 0 \\ \displaystyle2 \cos(θ) - 1=  0 \end{cases}

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

\begin{cases}\rm\displaystyle \displaystyle   \sin (θ)     \neq  - 2 \\ \displaystyle2 \cos(θ) =  1\end{cases}

divide both sides by 2:

\rm\displaystyle \displaystyle \displaystyle \cos(θ) =   \frac{1}{2}

by unit circle we get:

\rm\displaystyle \displaystyle \displaystyle θ=    {60}^{ \circ} , {300}^{ \circ}

so when θ is 60° a is:

\rm \displaystyle a =  \tan(  {60}^{ \circ} )  - 2 - 3 \sin(  {60}^{ \circ} )   +  2  \cos(  {60}^{ \circ} )

recall unit circle:

\rm \displaystyle a =   \sqrt{3}  - 2 -  \frac{ 3\sqrt{3} }{2}   +  2   \cdot  \frac{1}{2}

simplify which yields:

\rm \displaystyle a =    - \frac{   \sqrt{3} }{2}    - 1

when θ is 300°

\rm \displaystyle a =  \tan(  {300}^{ \circ} )  - 2 - 3 \sin(  {300}^{ \circ} )   +  2  \cos(  {300}^{ \circ} )

remember unit circle:

\rm \displaystyle a =  -  \sqrt{3}   - 2  +   \frac{3\sqrt{ 3} }{2}  +  2   \cdot  \frac{1}{2}

simplify which yields:

\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

5 0
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givi [52]

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Step-by-step explanation:

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