HELP ME PLS!!!!<br> WILL GIVE BRAINLIEST, 5 STARS, AND THANKS!!!

The figure shows a net of a cube. Which expression can be used to find the total surface area of the cube?

sergij07 [2.7K]

Answer:

Option C: 6(7 * 7)

Step-by-step explanation:

There are six faces of the cube. Since each face is 7 by 7 units, there are 6(7 *7) units total.

6 0

1 year ago

Read 2 more answers

Okay here is another math problem I'll give a brainliest for the right answer!

Sliva [168]

To find the total area of this figure, you have to find the area of each of the separate figures and then add those 2 answers together. Your work should look like this:

2 x 2 = 4

8 x 8 = 64

64 + 4 = 68

Your answer should be 68 ft.^2

4 0

3 years ago

if all of the medians of a triangle were constructed, what would the point where they all intersect be called?

Lubov Fominskaja [6]

Answer:

i think its centroid not sure though.

Step-by-step explanation:

3 0

3 years ago

Find the length of diagonal XU in the hexagon below. Round your solution to 2 decimal points

SpyIntel [72]

Answer:

The length of diagonal XU is 6.40 units

Step-by-step explanation:

* Lets explain how to find the distance between two points

- The rule of the distance between two points $(x_{1},y_{1})$

and $(x_{2},y_{2})$ is:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

* Lets solve the problem

-From the attached figure:

- The coordinates of the vertex X are (2 , 2)

- The coordinates of the vertex U are (-2 , 7)

∴ The point $(x_{1},y_{1})$ = (2 , 2)

∴ The point $(x_{2},y_{2})$ = (-2 , 7)

∴ $x_{1}=2,x_{2}=-2$

∴ $y_{1}=2,y_{2}=7$

∵ $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

∴ $d=\sqrt{(-2-2)^{2}+(7-2)^{2}}=\sqrt{(-4)^{2}+(5)^{2}}=\sqrt{16+25}=\sqrt{41}$

∵ $\sqrt{41}=6.403124$

∴ d = 6.40

* The length of diagonal XU is 6.40 units

6 0

3 years ago

A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximi

viktelen [127]

Answer:

Side length = $\sqrt{\frac{A}{3} }$ cm , Height = $\frac{1}{2} \sqrt{\frac{A}{3} }$ cm , Volume = $\frac{A\sqrt{A}}{6\sqrt{3} }$ cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

$y = \frac{A - x^{2} }{4x}$

Volume of box = Area x height

V = x² ₓ y

V = x² ₓ ( $\frac{A - x^{2} }{4x}$ )

V = $\frac{Ax - x^{3} }{4}$

To find max volume put V' = 0

So taking derivative equation becomes

$\frac{A - 3 x^{2} }{4} = 0$

A = 3 $x^{2}$

$x^{2}$ = $\frac{A}{3}$

x = $\sqrt{\frac{A}{3\\} }$

put value of x in equation 1

y = $\frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }$

y = $\frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }$

y = $\frac{1}{2} \sqrt{\frac{A}{3} }$

So the volume will be

V = $x^{2}$ × y

Put values of x and y from equation 2 & 3

V = $\frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )$

V = $\frac{A\sqrt{A}}{6\sqrt{3} }$

8 0

3 years ago

HELP ME PLS!!!! WILL GIVE BRAINLIEST, 5 STARS, AND THANKS!!!