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3 years ago

Write and Solve an area problem that illustrates the use of the find the pattern strategy

2 answers:

8 0

An area is when you multiply the height by the base. So, it depends on the picture and its numbers. ---------------------
For example: / 6in /
5in/ /
/ / model number: 5 in×6in=30 in
/ /
---------------------

Valentin [98]3 years ago

7 0

The answer above seems to be correct!

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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.

3 0

2 years ago

I need help sorry for bothering everyone. Two supporting reasons are missing from the proof. Complete the proof by dragging and

attashe74 [19]

Since ∠2=60° = ∠8 so it will also be 60 °
the sum of the both are 60°+∠7=180°
∠7= 180°- 60° = 120°

6 0

2 years ago

Read 2 more answers

Camison's minnan gets 14 miles per gallon for city driving and 19 miles per gan

loris [4]

Answer:

7 gallons were used for city driving and 17 gallons were used for highway driving

Correct statement and question:

camison's minnan gets 14 miles per gallon for city driving and 19 miles per gallon for highway driving. At the beginning of the week, the 24-gallon tank was full. The family traveled 421 miles before running out of gas. How many gallons were used for city driving and how many were used for highway driving?

Source:

Previous question found at brainly

Step-by-step explanation:

x = Gallons for city driving

24 - x = Gallons for highway driving

Miles drove before running out of gas = 421

Now let's solve for x, writing the following equation:

14x + 19 (24 - x) = 421

14x + 456 - 19x = 421

-5x = 421 - 456

-5x = - 35

x = -35/-5

x = 7 ⇒ 24 -x = 17

7 gallons were used for city driving and 17 gallons were used for highway driving

3 0

2 years ago

Identify the property used in the following a+b=b+a (ab)c =a(bc)

KATRIN_1 [288]

Answer: The first one is Commutative Property

The second one is associative property

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Solve this system of equations: 2x - y = 7 5x + y = 21

JulsSmile [24]

7x= 28 so x is 4 and y is 1.. So (4,1)

7 0

2 years ago