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3 years ago

Zach has a basic cell phone plan that does not include texting. He is going to add a multimedia texting package to his

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

Zach have to send 300 number of multimedia texts for the same cost in two packages.

Step-by-step explanation:

Package A charges $0.25 per multimedia text with no monthly fees.

Package B charges $0.20 per multimedia text and has a $15 monthly fee.

To frame the equations :

Let x be the any amount of multimedia texts are sent.
Let y be the total cost for using the multimedia package.

Package A Equation :

Total cost = No.of text sent × cost per text.

⇒ y = x × 0.25

⇒ y = 0.25x

∴ The equation of package A is y = 0.25x

Package B Equation :

Total cost = (No.of text sent × cost per text) + Monthly fee

⇒ y = (x × 0.20) + 15

⇒ y = 0.2x + 15

∴ The equation of a package B is y = 0.2x + 15

Now, the question is about how many multimedia texts will Zach have to send each month for the two multimedia texting packages to be the same cost.

⇒ equation of package A = equation of package B

⇒ 0.25x = 0.2x + 15

⇒ 0.25x - 0.2x = 15

⇒ 0.05x = 15

⇒ x = 15 / 0.05

⇒ x = 300

∴ Zach have to send 300 number of multimedia texts for the same cost in two packages.

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The function f(x)= -6x+11 has a range given by {-37,-25,-13,-1}.Select the domain values of the function from the list 1,2,3,4,5

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The range is {-37,-25,-13,-1}. So you need to figure out what four numbers from this list of numbers (1,2,3,4,5,6,7,8), when applied to this
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So you apply each of these numbers (1,2,3,4,5,6,7,8) into the function (f(x)=-6x+11)
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f(1)=-6(1)+11=5
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As you can see, f(2),f(4),f(6),and f(8) equal the numbers that are in the range {-37,-25,-13,-1}.

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A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

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Step-by-step explanation:

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And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is: