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3 years ago

Use the context of a 24-pack of soda to explain why 3/8 is NOT equivalent to 4/12.

Mathematics

1 answer:

Vsevolod [243]3 years ago

4 0

Answer:

Step-by-step explanation:

if we're going by a 24 pack of soda, 3/8 would be 9 sodas while 4/12 is a third, which would be 8 sodas

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For each of the following functions, determine if they are injective. Also determine if theyare surjective. Also determine if th

timurjin [86]

Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

$(a,b)=\left \{ x\in\Re : a$

2. Functions

1) $\Re \rightarrow \Re \\ f(x)=x$

Let's perform graph tests.

That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2) $\Re\geq0\rightarrow\Re , f(x)=x+1\\$

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3) $f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)$

The cosine function is not injective, bijective nor surjective.

4) $f:\Re\rightarrow\Re \:f(x)=ex$

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5) Quite unclear format

$6) \:f:\Re\rightarrow(0,\infty), f(x) =ex$

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

$7) f:\Re\geq 0 \rightarrow \Re\geq0, f(x) =x^{4}$

Not injective nor surjective therefore not bijective too.

$8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}$

$f(-1)=1, f(2)=4, f(-3)=9$

Injective (one to one), Surjective, and Bijective.

$10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.$

Surjective.

$11.f:R\geq 0[-1,1], f(x) = 0\\$

Surjective

12.f: US Citizens→Z, f(x) = the SSN of x.

General function

13.f: US Zip Codes→US States, f(x) = The state that x belongs to.

Surjective

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2 years ago

A group of health care workers were asked how many times per week they visited a gym. The data is shown on the dot plot. Select

mezya [45]

Answer:

pretty vague

Step-by-step explanation:

4 0

3 years ago

how do I solve this quadratic simultaneous equation 2y+2x=7 and x^2-4y^2=8 it has two sets of solutions and I'm solving by both

Ghella [55]

These are all the possible answers i know from experience.

2x + 2y = 7
x^2 - 4y^2 = 8

2x = -2y + 7
x = -y + 7/2

(-y + 7/2)^2 - 4y^2 = 8

y^2 - 7y + 49/4 - 4y^2 = 8

0 = 3y^2 + 7y + 8 - 49/4

0 = 12y^2 + 28y + 32 - 49

0 = 12y^2 + 28y - 17

y = (-b ± √(b^2 - 4ac))/(2a)

y = (-28 ± √(28^2 - 4(12)(-17)))/(2(12))

Dividing by 2 is like dividing by √(4):

y = (-14 ± √(7*28 + (12)(17)))/(2*6)

y = (-7 ± √(7*7 + (3)(17)))/6

y = (-7 ± 10)/6 = 1/2 or -17/6

x = -y + 7/2 = 3 or 19/3

<span>Solutions are (3,1/2) and (19/3,-17/6)</span>

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3 years ago

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Cloud [144]

Answer: 45.43%

Step-by-step explanation:

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