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Stells [14]
3 years ago
10

A penalty in Meteor - Mania is - 5 seconds. A penalty in Cosmic Calamity is - 7 seconds. Yolanda had penalties totaling -25 seco

nds in a game of meteor- Mania and -35 seconds in a game of Cosmic Calamity. In which game did Yolanda receive more penalties? Justify the answer.
Mathematics
1 answer:
Sonbull [250]3 years ago
3 0

Answer:

Yolanda had the same number of penalties in both games.

Step-by-step explanation:

Both of these penalties can be modeled by a first order equation.

Game of Meteor-Mania:

In a game of Meteor-Mania, each penalty is -5 seconds. So the expression for the total of penalties is:

Tp(n) = -5*n, where n is the number of penalties.

In the game of Meteor-Mania, Yolanda had penalties totaling -25 seconds. So

-25 = -5*n *(-1)

5n = 25

n = 25/5

n = 5

Yolanda had 5 penalties in the game of Meteor-Mania

Game of Cosmic Calamity

In a game of Meteor-Mania, each penalty is -7 seconds. So the expression for the total of penalties is:

Tp(n) = -7*n, where n is the number of penalties.

In the game of Cosmic Calamity, Yolanda had penalties totaling -35 seconds. So

-35 = -7n *(-1)

7n = 35

n = 35/7

n = 5

Yolanda had 5 penalties in the game of Cosmic Calamity

Yolanda had the same number of penalties in both games.

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the ratio of pine trees to elm trees in a park is 7:11. How many pine trees are there in the park if there are 68 more elm trees
Hitman42 [59]

Answer:

<em>The number of </em><em>pine tree is 119</em><em> and the number of </em><em>elm tree is 187.</em>

Step-by-step explanation:

The ratio of pine trees to elm trees in a park is 7:11

Let us assume that the number of pine tree is 7x and the number of elm tree is 11x.

It is also given that, there are 68 more elm trees as compared to the pine trees.

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8 0
3 years ago
Solve y=f(x) for x. Then find the input when the output is -3.
DochEvi [55]

Answer:

Please check the explanation

Step-by-step explanation:

Given the function

f\left(x\right)\:=\:\left(x-5\right)^3-1

Given that the output = -3

i.e. y = -3

now substituting the value y=-3 and solve for x to determine the input 'x'

\:\:y=\:\left(x-5\right)^3-1

-3\:=\:\left(x-5\right)^3-1\:\:\:

switch sides

\left(x-5\right)^3-1=-3

Add 1 to both sides

\left(x-5\right)^3-1+1=-3+1

\left(x-5\right)^3=-2

\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}

Thus, the input values are:

x=-\sqrt[3]{2}+5,\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}-i\frac{\sqrt[3]{2}\sqrt{3}}{2},\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}+i\frac{\sqrt[3]{2}\sqrt{3}}{2}

And the real input is:

x=-\sqrt[3]{2}+5

  • x=3.74
4 0
3 years ago
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