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2 years ago

5

Is Pluto still a planet

1 answer:

zubka84 [21]2 years ago

6 0

No it was not counted as a plant fir it is a dwarf planet

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what is the name of the islands west of ecuador that famed scientist charles darwin spent a lot of time studying?

Gwar [14]

Answer:

The Galapagos Islands were studied

8 0

1 year ago

Most people from Iran trace their origins back to what group?

storchak [24]

Answer:

It should be the persians.

Explanation:Technically it is persians since they're referred to that even today although iranians referred to themselves as aryans back then.

4 0

2 years ago

Brainiest for whoever answer correctly.

olga2289 [7]

Answer:

B. Death Valley, CA

Explanation:

Here are the elevations of each location:

Key West, Florida 18 feet

Death Valley, California -282.2 feet

Seattle, Washington 175 feet

Corpus Christi, Texas 7 feet

Because Death Valley sits at -282.2 feet below sea level, the lowest elevation, <u>B is the answer</u>.

Tip: Without knowing the exact elevations of each location, we could assume that a <em>valley</em> would be lowest.

8 0

2 years ago

The height of a right circular cone is multiplied by 3, but its radius remains fixed. By what factor will the volume of the cone

Orlov [11]

Answer:

A. 3

Explanation:

A cone is a three dimensional shape with a flat base and curved surface. A right circular cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base.

The volume of a right circular cone = πr²h/3; where r is the radius of the cone and h is the height of the cone.

Let us assume the cone has a height of h, and radius of r. Hence:

Volume (V₁) = πr²h/3

If the radius remains fixed and the height is multiplied by 3. Hence:

V₂ = πr²(3h)/3 = πr²h = 3V₁

The volume would be multiplied by 3

3 0

2 years ago

What is the value of X6? <br><br> Show the solution.

wariber [46]

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

2 years ago