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3 years ago

Someone talk to me bout anythang

Mathematics

1 answer:

enot [183]3 years ago

7 0

Answer:

do you think pineapple belongs on pizza ?

Step-by-step explanation:

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Find the coordinates of the midpoint of the segment with endpoints A(-5, 6) and B(3, 2). You must show all work in order to rece

OLga [1]

Midpoint Formula = $M = ( \frac{ x_{1}+x_{2}}{2}) + ( \frac{ y_{1} + y_{2}} {2})$

We are given the coordinates: A(-5,6) and B(3,2)
Plug the values into the equation to get:
M = ((3 - 5)/2, (2 + 6)/2)
Now, just solve it down to get M = (-1,4).

The midpoint of the segment with endpoints A(-5,6) and B (3,2) is (-1,4). Hope this helps ALot and have a wonderful day!

4 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

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

The marked are the trigonometric identities

3 0

2 years ago

What is 6.338 rounded to the nearest hundredth

azamat

Answer:

6.34

Step-by-step explanation:

brainliest pleassse

3 0

3 years ago

Please help me thank youuu

zhuklara [117]

Answer:

m<Q = 133°

Step-by-step explanation:

From the question given above, the following data were obtained:

m<P = (x + 13)°

m<Q = (10x + 13)°

m<R = (2x – 2)°

m<Q =?

Next, we shall determine the value of x. This can be obtained as follow:

m<P + m<Q + m<R = 180 (sum of angles in a triangle)

(x + 13)° + (10x + 13)° + (2x – 2)° = 180

x + 13 + 10x + 13 + 2x – 2 = 180

x + 10x + 2x + 13 + 13 – 2 = 180

13x + 24 = 180

Collect like terms

13x = 180 – 24

13x = 156

Divide both side by 13

x = 156 / 13

x = 12

Finally, we shall determine m<Q. This can be obtained as follow:

m<Q = (10x + 13)°

x = 12

m<Q = 10(12) + 13

m<Q = 120 + 13

m<Q = 133°

8 0

3 years ago

Make a graph of the following region 1< x+y in the standard (X

artcher [175]

Slope of 1 dotted line

4 0

4 years ago