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3 years ago

HELP

Mathematics

1 answer:

stira [4]3 years ago

5 0

You would just do 5.7-8.1, which is -3.6 degrees Celcius. Hope I helped!

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What are vertical angles always equal to

babymother [125]

Answer:

Vertical angles are always equal to one another. In general, we can say that, 2 pairs of vertical angles are formed when two lines intersect.

Step-by-step explanation:

Parallel lines are lines that do not meet at any point in a plane.

Intersecting lines are straight lines that meet or crosses each other at a certain point.

8 0

2 years ago

Plz need this asap ONLY ANSWER IF YOU KNOW IT will give brainliest

Natalija [7]

Answer:

<u>1</u> would be your y axis

Step-by-step explanation:

1. so the y axis is up and down

2. the x axis is horizontal side to side

3. the x axis is 0

4. the y axis is positive 1

5. So your anwser would be in the format (x,y)

6. It is (0,<u>1</u>)

7. It is positive

8. Plz name me brainliest I would really appreciate it if you need more help just ask me

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2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.