Give two different ratios with a description of the ratio relationship using the

Mathematics

2 answers:

Sonbull [250]3 years ago

8 0

3:7
since the amount of teachers are in multiples of 5, you can just divide both by 5 to get a new simplified ratio

Shalnov [3]3 years ago

8 0

This has nothing to do with the question but with the adds it makes you watch you can skip them by putting your finger on the bar at the bottom and sliding your finger right :)

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Simplify the product using distributive property 5(x+3) show work

GrogVix [38]

Answer:

Step-by-step explanation:

5(x + 3) Remove the brackets.

5*x + 3*5 Combine

5x + 15

4 0

3 years ago

Simplify 2a^2b^3(4a^2+3ab^2-ab)

-BARSIC- [3]

2a^2b^3(4a^2+3ab^2-ab)=?

is what I presume you actually meant. 

Pull out the common factors of (4a^2+3ab^2-ab) and you will get 

a(4a+3b^2 -b) 

Substitute this back into the original equation and you get

2a^2b^3[a(4a+3b^2-b)] = 
2a^3b^3(4a+3b^2-b) =
2a^3b^3(4a-b+3b^2)

3 0

3 years ago

What’s the domain and range of: log(√(2x-1) + 3 ) Please explain how you got it too!!

Radda [10]

Two main facts are needed here:

1. The logarithm $\log x$ , regardless of the base of the logarithm, exists for $x>0$ .

2. The square root $\sqrt x$ exists for $x\ge0$ .

(in both cases we're assuming real-valued functions only)

By (2) we know that $\sqrt{2x-1}$ exists if $2x-1\ge0$ , or $x\ge\dfrac12$ .

By (1), we know that $\log(\sqrt{2x-1}+3)$ exists if $\sqrt{2x-1}+3>0$ , or $\sqrt{2x-1}>-3$ . But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require $x\ge\dfrac12$ , so the function has domain $\left[\dfrac12,\infty)$ .

To determine the range, we need to know that, in their respective domains, $\sqrt x$ and $\log x$ increase monotonically without bound. We also know that $x=\dfrac12$ at minimum, at which point the square root term vanishes, so the least value the function takes on is $\log3$ . Then its range would be $[\log3,\infty)$ .