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3 years ago

What is the height of a cone with a radius of 2 inches and volume of 6 cubic inches? Round your answer to the tenths place.

Mathematics

2 answers:

kondor19780726 [428]3 years ago

8 0

Answer:

1.4

Step-by-step explanation:

got it right on test

kykrilka [37]3 years ago

5 0

I hope this helps you

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I think of a number,divide it by 4 then subtract 3. Give your answer in algebraic notation.

Vilka [71]

X/4 - 3

(*make sure to write the division in fraction form as shown for algebraic notation)

5 0

2 years ago

Factor 27 + d^ 3completely.

Sedaia [141]

Answer:

(3+d)(9-3d+d^2)

Step-by-step explanation:

a^3 + b^3 = (a+b)(a^2-ab+b^2)

a= 3

b= d

7 0

3 years ago

A coffee shop covers it coffee cups for holiday seasons. Their cups are cylindrical, have a 2 inch radius, and a height of 6 inc

slega [8]

Answer:

Surface area of square box cover is 4x12=48 $in^{2}$

Step-by-step explanation:

Given cylindrical cup have 2 in of radius and height of 6 in.

It is said that we need to cover the cup leaving top and bottom of cup.

To find how much material to cover:

After covering the cup, square box will be formed around the cup.

As shown in figure, Top view of square cup and box.

One face of box has length of diameter of cup = 2 in and height of box as height of cup = 6 in

Therefore, Area of one face of square box is length x height = 2 x 6 =12 $in^{2}$

Since, Box having 4 faces

Surface area of square box cover = 4 x area of one face of box

= 4x12=48 $in^{2}$

8 0

3 years ago

please help me, Prove a quadrilateral with vertices G(1,-1), H(5,1), I(4,3) and J(0,1) is a rectangle using the parallelogram me

mestny [16]

Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are equal, thus

Join JH and GI such that they form the diagonals of the quadrilateral.Now,

JH= $\sqrt{(5-0)^{2}+(1-1)^{2}}=5$ and

GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$

Now, mid point of JH= $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

= $(\frac{5+0}{2},\frac{1+1}{2})=(\frac{5}{2},1)$

Mid point of GI= $(\frac{5}{2},1)$

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

$(GI)^{2}=(IJ)^{2}+(JG)^{2}$ (1)

Now, JI= $\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20}$ and GJ= $\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}$

Substituting these values in (1), we get

$5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }$

$25=20+5$

$25=25$

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$ and HJ= $\sqrt{(0-5)^2+(1-1)^2}=5$ are equal, thus, GHIJ is a rectangle.

6 0

3 years ago

Find the area of the figure below

tatuchka [14]

Step-by-step explanation:

(3×6) + (2×6)

= 18 + 12

= 30 in²

4 0

2 years ago