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3 years ago

What is the area of this parallelogram?

Mathematics

1 answer:

Alex Ar [27]3 years ago

7 0

Formula for area of parallelogram:
A= bh
Where b is the length of base and h is the height.
Here b=2.3+4.5 = 6.8 cm
and h= 4.5 cm
Put values in formula.

A= 6.8 x 4.5
A=30.6 cm²

Answer: 30.6 cm²

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Five kittens are sharing 6 cups of milk equally. How much milk does each kitten get?

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2 years ago

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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.

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2 years ago

A hawk is sitting on the branch of a tree Question Image ft above the ground. It flies 28.3 ft higher in the air, drops down by

nikklg [1K]

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How to solve this problem?

Mrrafil [7]

10.9297 anxdjehsimdudneueneieo

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2 years ago

Use the parabola tool to graph the quadratic function y=-1? - 2.0 +8

Andreas93 [3]

Answer:

see explanation

Step-by-step explanation:

I don't have graphing facilities but can give you the vertex and 1 other point.

Given a parabola in standard form

y = ax² + bx + c ( a ≠ 0 )

Then the x- coordinate of the vertex is

x = - $\frac{b}{2a}$

y = - x² - 2x + 8 ← is in standard form

with a = - 1 and b = - 2 , then

x = - $\frac{-2}{-2}$ = - 1

Substitute x = - 1 into the equation for corresponding value of y

y = - (- 1)² - 2(- 1) + 8 = - 1 + 2 + 8 = 9

vertex = (- 1, 9 )

To obtain another point substitute any value for x into the equation

x = 0 : y = 0 - 0 + 8 , then (0, 8 ) is a point on the graph

x = 2 : y = - (2)² - 2(2) + 8 = - 4 - 4 + 8 = 0 then (2, 0 ) is a point on the graph

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3 years ago