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1 year ago

If a container has 5 rows of boxes that are stacked 3 boxes high and 2 boxes deep, how many boxes are in the container?

Mathematics

2 answers:

aleksandrvk [35]1 year ago

8 0

<h2>Answer:</h2>

The total number of boxes that are in the container are:

30 boxes.

<h2>Step-by-step explanation:</h2>

It is given that:

A container has 5 rows of boxes that are stacked 3 boxes high and 2 boxes deep.

This means that the total number of boxes is calculated as the product of number of boxes in a row multiply by the number of boxes along a column and the boxes above other i.e. the number of boxes in depth.

Hence, the number of boxes are: 5×3×2=30 boxes.

kaheart [24]1 year ago

3 0

There are 30 boxes in the container

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2 years ago

R(-9, 4) and S(2,-1); Find T.

madreJ [45]

Answer:

12.083 units

Step-by-step explanation:

Here we are given with two coordinates and asked to determine the distance between them.

Here we are going to use the distance formula, which is given as under

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Where

$(x_1,y_1) = (-9,4)$

$(x_2,y_2) = (2,-1)$

Replacing these values in the distance formula

$D =\sqrt{(-9-2))^2+(4-(-1))^2}$

$D =\sqrt{(-11)^2+(5)^2}$

$D =\sqrt{121+25}$

$D =\sqrt{146}$

$D= 12.083$

Hence the distance is 12.083 units

3 0

1 year ago

What is the binomial expansion of (x + 2)4? x4 + 4x3 + 6x2 + 4x + 1 8x3 + 24x2 + 32x x4 + 8x3 + 24x2 + 32x + 16 2x4 + 8x3 + 12x2

Margaret [11]

<u>Answer-</u>

$\boxed{\boxed{(x+2)^4=x^4+8x^3+24x^2+32x+16}}$

<u>Solution-</u>

Given expression is $(x+2)^4$

Applying Binomial Theorem

$\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

Here,

a = x, b = 2 and n = 4

So,

$\left(x+2\right)^4=\sum _{i=0}^4\binom{4}{i}x^{\left(4-i\right)}\cdot \:2^i$

Expanding the summation

$=\dfrac{4!}{0!\left(4-0\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(4-1\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(4-2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(4-3\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(4-4\right)!}x^0\cdot \:2^4$

$=\dfrac{4!}{0!\left(4\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(3\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(1\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(0\right)!}x^0\cdot \:2^4$