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dexar [7]
2 years ago
11

There are 16 types of flowers used to decorate for a party. Twelve of the flowers types last an average of 4 days before they wi

lt. The remaining flowers last an average of 6 days.
What is the average number of days before the flowers wilt?

PLEASE HELP ASAP THANKS SO MUCH!!!! :))))
Mathematics
1 answer:
Bumek [7]2 years ago
6 0
16 - 12 = 4 flowers
4f = 6 days
1 flower = 1.5 days

1.5 × 16 = 24 days



Hope this helped!!
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Answer:

The variable c represents the domain as it is the independent variable.

The domain of the function F(c) is given by c ≥ 0.

So, only positive values for the input make sense.

The upper limit of the domain is +∞ and lower limit is 0.

It is not possible for the team to earn $50.50 as it will be only multiple of 2.

Step-by-step explanation:

If F(c) represents the earning of a volleyball team from selling c cupcakes and each cupcake costs $2 each, then the equation that models the situation is  

F(c) = 2c ..... (1)

The variable c represents the domain as it is the independent variable. (Answer)

The domain of the function F(c) is given by c ≥ 0. (Answer)

So, only positive values for the input make sense. (Answer)

The upper limit of the domain is +∞ and the lower limit is 0. (Answer)

It is not possible for the team to earn $50.50 as it will be only multiple of 2. (Answer)

5 0
2 years ago
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A botanist is using two types of plants for an experiment. She writes inequalities to model the constraints on the number of eac
Valentin [98]
The vertex (5,39)

5 is the value of x. 39 is the value of y. y is the cost function of the minimum value in dollars.

(5,39) vertex means that  <span>Buying five of each type of plant costs $39, which is the lowest possible cost.</span>
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S = √3*73.5 /4.5=7 (Khufu’s pyramid)
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√3*<span>9/10 </span>/4.5=0.734 i hope this  helps you
4 0
2 years ago
What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove
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Answer:

The solutions are:

b=0,\:b=4

Step-by-step explanation:

Considering the expression

  • \frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

Solving the expression

\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2

5b^3-10=6b^3-4b^2-10

\mathrm{Switch\:sides}

6b^3-4b^2-10=5b^3-10

6b^3-4b^2-10+10=5b^3-10+10

6b^3-4b^2=5b^3

\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}

6b^3-4b^2-5b^3=5b^3-5b^3

b^3-4b^2=0

Using\:the\:Zero\:Factor\:Principle: if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

So,

b=0,b-4=0

b=0,b=4

Therefore, the solutions are:

b=0,\:b=4

4 0
2 years ago
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loris [4]

Here we are given with a triangle with smaller triangles formed due to the altitude on AC. Given:

  • AB = 6
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We have to find the value for sin \theta

So, Let's start solving....

In ∆ADB and ∆ABC,

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So, ∆ADB ~ ∆ABC (By AA similarity)

The corresponding sides will be:

\sf{ \dfrac{AD}{AB}  =  \dfrac{AB}{AC} }

We know the value of AB and to find AC, we can use Pythagoras theoram that is:

AC = √6² + 8²

AC = 10

Coming back to the relation,

\sf{ \dfrac{AD}{6}  =  \dfrac{6}{10} }

\sf{AD =  \dfrac{6 \times 6}{10}  = 3.6}

In ∆ADB, we have to find sin \theta which is given by perpendicular/base:

\sf{\sin( \theta)  =  \dfrac{AD}{AB} }

Plugging the values of AD and AB,

\sf{\sin( \theta)  =  \dfrac{3.6}{6} }

Simplifying,

\sf{ \sin( \theta)  =  \dfrac{3}{5}  =  \boxed{ \red{0.6}}}

And this is our final answer.....

Carry On Learning !

4 0
2 years ago
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