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Liono4ka [1.6K]
3 years ago
14

What is the slope of a line segment with endpoints at (-1,1) and (1,5)?

Mathematics
1 answer:
torisob [31]3 years ago
4 0
The answer is 2 use the x1 x2 over y1 y2 formula
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Find the area of the figure.( sides meet at right angles.)picture below
Tema [17]

You can divide the figure as follows:

  • A vertical rectangle, 5 meters tall and 2 meters wide, on the top right.
  • A horizontal rectangle, 2 meters tall and 5 meters wide, on the bottom left
  • A 2x2 meters square on the bottom right, where the two rectangles meet.

The area of the rectangles is 5x2=10 meters squared, while the area of the square is 2x2=4 meters squared.

So, the total area is 10+4=14 meters squared.

5 0
4 years ago
Root 3 cosec140° - sec140°=4<br>prove that<br><br>​
Lorico [155]

Answer:

Step-by-step explanation:

We are to show that \sqrt{3} cosec140^{0} - sec140^{0} = 4\\

<u>Proof:</u>

From trigonometry identity;

cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}

\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\

From trigonometry, 2sinAcosA = Sin2A

= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\=  \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\=  \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2  \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

\frac{4(sin(420-140)) }{sin280}

= \frac{4sin280}{sin280}\\ = 4 \ Proved!

3 0
3 years ago
The diagonals of kite KITE intersect at point P. If m A. 34°<br> B. 46°<br> C. 68°<br> D. 92°
JulijaS [17]

Answer:

The diagonal of kite intersect at the point p is C. 68°

7 0
3 years ago
Use the properties of exponents to write an equivalent expression that is a product of unique
jolli1 [7]

Ummm I just need to answer questions sorry!!!

5 0
3 years ago
HOW do you find the value of TAN 88 degrees 22' 45'' and COT 36 degrees???
storchak [24]
Part 1
We want to find the tan of 88 degrees, 22' 45''.

88° 22' 45''
= 88 + 22/60 + 45/3600 degrees
= 88.3792°

From the calculator, obtain
tan(88.3792°) = 35.3409

Answer: 35.3409

Part 2
We want to find cot(36°).

By definition, cot(x) = 1/tan(x).
From the calculator, obtain
tan(36°) = 0.7265
Therefore
cot(36°) = 1/0.7265 = 1.3764

Answer: 1.3764

3 0
4 years ago
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