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Contact [7]
1 year ago
9

What are the 7 classifications of oak tree?

Geography
1 answer:
Leni [432]1 year ago
4 0

The 7 classifications of oak tree:

Black Oak (Quercus Velutina) .

Northern Red Oak Tree (Quercus rubra) .

White Oak Tree (Quercus alba) .

Scarlet Oak (Quercus coccinea) .

Southern Red Oak (Quercus falcata) .

Post Oak (Quercus Stellata) .

Chestnut Oak (Quercus Prinus)

Alright wooden has a density of zero. seventy five grams in step with cubic centimetre, which is extra than pine wood that carries 0.forty three grams consistent with cubic Centimetre . consequently, okay wooden is seemed because the maximum long lasting, difficult and fungal-resistant cloth, specially used for making ecu alright furniture and doorways.

The historic Hebrews considered the all Rightsacred as it became beneath an alrightthat Abraham gave hospitality to God and two of his angels, who had been disguised as guests. This story is advised in Genesis 18; and is certainly one of 60 references to o.k.in the Bible. The early Gauls worshipped very wellas a image of their ideal God.

Learn more about oak tree: here:

brainly.com/question/21372028

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What is the value of X6? <br><br> Show the solution.
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Answer : The value of x_6 is \sqrt{7}.

Explanation :

As we are given 6 right angled triangle in the given figure.

First we have to calculate the value of x_1.

Using Pythagoras theorem in triangle 1 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_1)^2=(1)^2+(1)^2

x_1=\sqrt{(1)^2+(1)^2}

x_1=\sqrt{2}

Now we have to calculate the value of x_2.

Using Pythagoras theorem in triangle 2 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_2)^2=(1)^2+(X_1)^2

(x_2)^2=(1)^2+(\sqrt{2})^2

x_2=\sqrt{(1)^2+(\sqrt{2})^2}

x_2=\sqrt{3}

Now we have to calculate the value of x_3.

Using Pythagoras theorem in triangle 3 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_3)^2=(1)^2+(X_2)^2

(x_3)^2=(1)^2+(\sqrt{3})^2

x_3=\sqrt{(1)^2+(\sqrt{3})^2}

x_3=\sqrt{4}

Now we have to calculate the value of x_4.

Using Pythagoras theorem in triangle 4 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_4)^2=(1)^2+(X_3)^2

(x_4)^2=(1)^2+(\sqrt{4})^2

x_4=\sqrt{(1)^2+(\sqrt{4})^2}

x_4=\sqrt{5}

Now we have to calculate the value of x_5.

Using Pythagoras theorem in triangle 5 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_5)^2=(1)^2+(X_4)^2

(x_5)^2=(1)^2+(\sqrt{5})^2

x_5=\sqrt{(1)^2+(\sqrt{5})^2}

x_5=\sqrt{6}

Now we have to calculate the value of x_6.

Using Pythagoras theorem in triangle 6 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_6)^2=(1)^2+(X_5)^2

(x_6)^2=(1)^2+(\sqrt{6})^2

x_6=\sqrt{(1)^2+(\sqrt{6})^2}

x_6=\sqrt{7}

Therefore, the value of x_6 is \sqrt{7}.

5 0
3 years ago
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