All categories

2 years ago

What is the inverse of -2

Mathematics

1 answer:

Aleksandr [31]2 years ago

6 0

Solution:

The multiplicative inverse of (-2) is -1/2 {Reciprocal}.

The additive inverse of (-2) is 2.

You might be interested in

Find the solution of the given initial value problem. ty' + 6y = t2 − t + 1, y(1) = 1 6 , t > 0

gtnhenbr [62]

Answer:

Step-by-step explanation:

Given is a differential equation as

$ty' + 6y = t^2 - t + 1, y(1) = 1 6 , t > 0$

Divide this by t to get in linear form

$y'+6y/t = t-1+1/t$

This is of the form

y' +p(t) y = Q(t)

where p(t) = 1/t $e^(\int 1/tdt) = t$

So solution would be

$yt = \int t^2-t+1 dt\\= t^3/3-t^2/2+t+C$

siubstitute y(1) = 16

$16 = 16^3/3-128+1+C\\C = -1206$

4 0

2 years ago

Please help i don’t understand

Anit [1.1K]

A line is 180 so you subtract 126 with 180 and get 54. Since a Triangle angles all equal 180 together you add 54 and 43 and subtract 180 and the answer is 83

4 0

2 years ago

Read 2 more answers

Simple interest 34,100 at 4% for 3 years

Kaylis [27]

Answer:

Step-by-step explanation:

.....

4 0

2 years ago

Read 2 more answers

Mr. Brady is using a coordinate plane to design a treasure hunt for his students. The hunt begins at the flagpole. The first clu

dmitriy555 [2]

Answer:

The figure is attached and the total distance is 1031 feet.

Step-by-step explanation:

The graph is indicated in the attached figure.

For calculation of distance consider following

Point Flagpole is (0,0)

Point Clue1 is (0,5)

Point Clue2 is (6,0)

Point Clue3 is (0,-5)

So the distance is calculated as follows

$d_{T}=[d_{FP\ to\ Clue1}+d_{Clue1\ to\ Clue2}+d_{Clue2\ to\ Clue3}]*distance\ per\ unit\\d_{T}=[\sqrt{(Clue1_x-FP_x)^2+(Clue1_y-FP_y)^2}+\sqrt{(Clue2_x-Clue1_x)^2+(Clue2_y-Clue1_y)^2}+\sqrt{(Clue3_x-Clue2_x)^2+(Clue3_y-Clue2_y)^2}]**distance\ per\ unit\\$ Substituting the values

$d_{T}=[\sqrt{(0-0)^2+(5-0)^2}+\sqrt{(6-0)^2+(0-5)^2}+\sqrt{(0-6)^2+(-5-0)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{(0)^2+(5)^2}+\sqrt{(6)^2+(-5)^2}+\sqrt{(-6)^2+(-5)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{0+25}+\sqrt{36+25}+\sqrt{36+25}]*50 \text{ feet}\\d_{T}=[\sqrt{25}+\sqrt{61}+\sqrt{61}]*50 \text{ feet}\\d_{T}=[5+7.81+7.81]*50 \text{ feet}\\d_{T}=[20.62]*50 \text{ feet}\\d_{T}=1031 \text{ feet}\\$