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3 years ago

Help me on number two plz.

Mathematics

1 answer:

SpyIntel [72]3 years ago

5 0

The Answer Here, Is 33. -8 Is 8 From 0. And 25 Is 25 from 0. 25 + 8 = 33 So Your Answer Is 23

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Point (, ) is dilated from the origin by scale factor = . What are the coordinates of point ′?

Igoryamba

Answer:idk how to do this im in 7th grade

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3 years ago

The table below shows field goal statistics for three different kickers

FromTheMoon [43]

Answer:

<h2>From least to greatest: C, B and A</h2>

Step-by-step explanation:

To order each Kicker correctly, we have to find the success rate of each one, which is obtained by dividing and then multiplying by 100 to obtained the percentage:

$R=\frac{made}{attempted}$

<h3>Success rate of A:</h3>

$R_{A}=\frac{12}{14}.100=86\%$

<h3>Success rate of B:</h3>

$R_{B}=\frac{15}{18}.100=83\%$

<h3>Success rate of C:</h3>

$R_{C}=\frac{19}{24}.100=79\%$

So, comparing all three, from least to greatest we have:

C with 79%, then B with 83%, and A with 86%.

3 0

3 years ago

Read 2 more answers

If you throw a ball at a rectangular board that is 4 ft. by 5 ft. with a smaller 2 ft. by 2 ft. square painted in the center of

bekas [8.4K]

Answer:

Step-by-step explanation:

Probability=(2*2)/(4*5)=1/5=20 %

8 0

2 years ago

The midpoint of AB is M(0, 3) If the coordinates of A are (4, - 1) , what are the coordinates of B?

Harlamova29_29 [7]

-4;4

XB=0*2-4

YB=3*2-(-1)

3 0

2 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

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

The marked are the trigonometric identities

3 0

1 year ago