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Butoxors [25]
3 years ago
12

A skier is trying to decide whether or not to buy a season ski pass. A daily pass costs ​$68. A season ski pass costs ​$300. The

skier would have to rent skis with either pass for ​$20 per day. How many days would the skier have to go skiing in order to make the season pass less expensive than the daily​ passes?
Mathematics
1 answer:
lina2011 [118]3 years ago
5 0

first, ignore the rent for the skies. we could add it to both, but since it's the same, it would not change the number of days here.

this information is meant to test your decision making on what's important for the question.

now just ask how often you can pay $68 dollars per day while staying below 300 in total, you can just count it up.

68, 136, 204, 272, 340

its 4 days

for 5 days or more, the season pass is the better option

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Difference of Squares gives which complex factors for the expression x2 +3?
Andreas93 [3]

Step-by-step explanation:

Shortest way to solve this question is to find the factors of the given expression.

The given expression is (x² + 13).

Now we have to factorize it.

(x² + 13) = x² + (√13)²

            = x² + [-(i)²√(13)²]  [Since i = √(-1)]

            = x² - (i√13)²

            = (x - i√3)(x + i√3) [Since (a² - b²) = (a + b)(a - b)]-by-

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3 years ago
Will give brainliest, thanks, and 5 stars! LOTS OF POINTS! 50 POINTS!
sweet-ann [11.9K]

Answer/Step-by-step explanation:

 All of the solutions to the equation 3x^2 - 12 = 0 are x = 12 and x = -2

Answer: False

Explanation:

3x^2-12+12=0+12

3x^2=12

\frac{3x^2}{3}=\frac{12}{3}

x^2=4

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{4},\:x=-\sqrt{4}

x=2,\:x=-2

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There are two unique solutions to the equations (x-3)^2 = 16

Note: Each variable in the matrix can have only one possible value, and this is how you know that this matrix has one unique solution.

Answer: True

Explanation:

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=7,\:x=-1

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Ths solutions for the equation 2(x-3)^3 - 18 = 0 are x = 6 and x = 0

Answer: False

Explanation:

2\left(x-3\right)^3-18+18=0+18

2\left(x-3\right)^3=18

\frac{2\left(x-3\right)^3}{2}=\frac{18}{2}

\left(x-3\right)^3=9

\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}

=\sqrt[3]{9}+3,\:x=\frac{6-\sqrt[3]{9}}{2}+i\frac{\sqrt[3]{9}\sqrt{3}}{2},\:x=\frac{6-\sqrt[3]{9}}{2}-i\frac{\sqrt[3]{9}\sqrt{3}}{2}

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The solutions for the equation 2(x-5)^2-8=0 are x = 7 and x = -7

Answer: False

Explanation:

2\left(x-5\right)^2-8+8=0+8

2\left(x-5\right)^2=8

\frac{2\left(x-5\right)^2}{2}=\frac{8}{2}

\left(x-5\right)^2=4

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=7,\:x=3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation (x + 3)^2-25 = -8 are x = 2 and x = -8

Answer: False

Explanation:

\left(x+3\right)^2-25+25=-8+25

\left(x+3\right)^2=17

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{17}-3,\:x=-\sqrt{17}-3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 2(2x-1)^2=18 are x = 5 and x = -4

Answer: False

Explanation:

\frac{2\left(2x-1\right)^2}{2}=\frac{18}{2}

\left(2x-1\right)^2=9

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=2,x=-1

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The only solution for equation (2x-1)^2-49=0 is x = 4

Answer: False
Explanation:

\left(2x-1\right)^2-49+49=0+49

\left(2x-1\right)^2=49

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=4,\:x=-3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 3(x+2)^2 - 3 = 0 are x = -3 and x = -1

Answer: True
Explanation:

3\left(x+2\right)^2-3+3=0+3

3\left(x+2\right)^2=3

\frac{3\left(x+2\right)^2}{3}=\frac{3}{3}

\left(x+2\right)^2=1

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=-1,\:x=-3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 5x^2 - 180 = 0 are x = 6 and x  = -6

Answer: True

Explanation:

5x^2-180+180=0+180

5x^2=180

\frac{5x^2}{5}=\frac{180}{5}

x^2=36

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{36},\:x=-\sqrt{36}

x=6,\:x=-6

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