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3 years ago

Calculate the surface area of a solid box in the shape of a cube with side length 7cm.

Mathematics

2 answers:

goldfiish [28.3K]3 years ago

8 0

Step-by-step explanation:

a=7

=6a^2

=6×(7)^2

=6×49

=294

melamori03 [73]3 years ago

5 0

Answer:

Surface area of the cube = 294cm^2

Step-by-step explanation:

A cube has the same side lengths for every side

Surface area of one face is length * length

7 * 7 = 49cm^2

A cube has 6 faces

49 * 6 = 294cm^2

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What is the measure of angle z in this figure?

astraxan [27]

Hello!

I've attached a photo for reference.

Lines A and B form straight angles, which measure 180 degrees. That means that -

m∠x + m∠y = 180°
m∠y + m∠z = 180°
m∠z + 43° = 180°
43° + m∠x = 180°

Since you're trying to find z, use the solvable equation with z in it:

m∠z + 43° = 180°
180 = z + 43
137 = z

Answer:
m∠z = 137°

3 0

3 years ago

Three-fifths of the students are girls. One-third of the girls have blond hair. One-half of the boys have brown hair. A:what fra

jeka94

The answer is 1/5 of the students are blond girls, since 1/3 of 3/5 is 1/5.
Divide 3/5 by 3, which is 1/5.
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5 0

3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$