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2 years ago

Help me please ! Also dont be offended if i message back saying why it might be wrong

Mathematics

1 answer:

Drupady [299]2 years ago

6 0

Answer:

$1mm = 0.001m \\ 1000mm = 1m \\ 1cm = 0.01m \\ 100cm = 1m \\ 1km = 1000m \\ 0.001km = 1m \\ 1cm = 10mm \\ 0.1cm = 1mm \\ \\ m \: \: meter \\ cm \: \: centimeter \\ mm \: \: millimeter \\ km \: \: kilometer$

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A university is building a new student center that is two- thirds the distance from the arts center to the residential complex.

Natasha2012 [34]

Answer:

$C = (\frac{21}{5},\frac{33}{5})$

Step-by-step explanation:

Given

Points: (1, 9) and (9, 3)

Ratio = 2/3

Required

Determine the coordinate of the center

Represent the ratio as ratio

$Ratio = 2:3$

The new coordinate can be calculated using

$C = (\frac{mx_2 + nx_1}{n + m},\frac{my_2 + ny_1}{n + m})$

Where

$(x_1,y_1) = (1, 9)$

$(x_2, y_2) = (9, 3)$

$m:n = 2:3$

Substitute these values in the equation above

$C = (\frac{2 * 9 + 3 * 1}{3 + 2},\frac{2 * 3 + 3 * 9}{2 + 3})$

$C = (\frac{18 + 3}{5},\frac{6 + 27}{5})$

$C = (\frac{21}{5},\frac{33}{5})$

Hence;

<em>The coordinates of the new center is </em> $C = (\frac{21}{5},\frac{33}{5})$ <em></em>

7 0

3 years ago

sophia earned $52 at her job when she workd for 2 hours. what was her hourly pay rate in hours per dollar?

statuscvo [17]

1 hour =$26, she has an hourly rate of 26 dollars

3 0

3 years ago

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities