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3 years ago

Highlight very briefly why/how when you subtract a negative number, that it yields a positive number. Will give brainliest

Mathematics

1 answer:

lbvjy [14]3 years ago

3 0

When we build integers from natural numbers, we're looking for additive inverse of natural numbers?

What's an additive inverse? Well, for example, the additive inverse of 2 is a numbers $x$ such that

$2+x = 0$

We call this number -2. So, the real meaning behind the negative sign is "if you add me and my positive counterpart, the result is zero".

So, -5 is the additive inverse of 5, -16 is the additive inverse of 16, and so on, because

$5-5=0,\quad 16-16=0,\ldots$

Note that this is a symmetrical relation: if -5 is the inverse of 5, it is also true that 5 is the inverse of -5.

So, when you write something like

$5-(-4)$

it means that you want to add 5 and the inverse of the inverse of 4. But given what we just said, the inverse of the inverse of a number is the number itself, which is why subtracting a negative number is the same as adding it.

You might be interested in

A given line has the equation 10x + 2y = −2.

worty [1.4K]

The equation is $\boxed{ \ y = - 5x + 12 \ or \ y = 12 - 5x} \ }$

<h3>Further explanation </h3>

This case asking the end result in the form of a slope-intercept.

<u>Step-1: find out the gradient. </u>

10x + 2y = -2

We isolate the y variable on the left side. Subtract both sides by 10x, we get:

2y = - 10x - 2

Divide both sides by two

y = -5x -1

The slope-intercept form is $\boxed{ \ y = mx + c \ }$ , with the coefficient m as a gradient. Therefore, the gradient is m = -5.

If you want a shortcut to find a gradient from the standard form, implement this:

$\boxed{ \ ax + by = k \rightarrow m = - \frac{a}{b} \ }$

10x + 2y = −2 ⇒ a = 10, b = 2

$\boxed{m = - \frac{10}{2} \rightarrow m = -5}$

<u>Step-2:</u> the conditions of the two parallel lines

The gradient of parallel lines is the same $\boxed{ \ m_1 = m_2 \ }$ . So $\boxed{m_1 = m_2 = -5}.$

<u>Final step:</u> figure out the equation, in slope-intercept form, of the parallel line to the given line and passes through the point (0, 12)

We use the point-slope form.

$\boxed{ \ \boxed{ \ y - y_1 = m(x - x_1)} \ }$

Given that

m = -5
(x₁, y₁) = (0, 12)

y - 12 = - 5(x - 0)

y - 12 = - 5x

After adding both sides by 12, the results is $\boxed{ \ y = - 5x + 12 \ or \ y = 12 - 5x} \ }$

<u>Alternative steps </u>

Substitutes m = -5 and (0, 12) to slope-intercept form $\boxed{ \ y = mx + c \ }$

12 = -5(0) + c

Constant c is 12 then arrange the slope-intercept form.

Similar results as above, i.e. $\boxed{ \ y = - 5x + 12 \ or \ y = 12 - 5x} \ }$

$\boxed{Standard \ form: ax + by = c, with \ a > 0}$

$\boxed{Point-slope \ form: y - y_1 = m(x - x_1)}$

$\boxed{Slope-intercept \ form: y = mx + k}$

<h3>Learn more </h3>

A similar problem brainly.com/question/10704388
Investigate the relationship between two lines brainly.com/question/3238013
Write the line equation from the graph brainly.com/question/2564656

Keywords: given line, the equation, slope-intercept form, standard form, point-slope, gradien, parallel, perpendicular, passes, through the point, constant

7 0

3 years ago

Read 2 more answers

Young people often find that playing video games, watching TV, and instant messaging friends can be very relaxing, and can help

sergeinik [125]

Answer:

The probability that a student in this survey says something other than that he or she needs a vacation is:

= 58%.

Step-by-step explanation:

The probability of the teens at the local high school in Oregon who said that they needed a vacation = 42%,

Therefore, the probability that a student in the same survey says something other than that he or she needs a vacation must be 58% (100% - 42%).

Probability calculates the frequency of the occurrences of an event.

3 0

3 years ago

Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$