A train leaves the station at time t = 0. Traveling at a constant speed, the train travels 390

Mathematics

1 answer:

maxonik [38]3 years ago

8 0

Answer:

a: D= 130T

B it is a linear function

Step-by-step explanation:

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A company has budgeted 5 1/4 hours to complete a project, with 1/3 of the time spent on research. How much time does the company

elena-s [515]

Given that a company budgeted 5 1/4 hours to complete a project, determine how much time they spent on research if they spent 1/3 of the total budget.

First, convert the budget from hours into minutes.

5 1/4 hours = 315 minutes

1 hour = 60 min

1/4 hour = 15 min

60 x 5 = 300

300 + 15 = 315

Then, divide the minutes by 3 or multiply it by 1/3.

315 / 3 = 105

315 x 1/3 = 105

Lastly convert to a mixed number.

1 3/4 hour

Thus, the company plans to spend 1 3/4 hours or 1 hour and 45 minutes on research.

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3 years ago

I Need help again :)

Nookie1986 [14]

B it’s b I think I did same one

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2 years ago

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Solve the following equation by completing the square. 3x^2-3x-5=13

mr Goodwill [35]

we'll start off by grouping some

$\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6$

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

$\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2$

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

$\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}$

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

$\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}$