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3 years ago

Which two figures have the same number of faces?

Mathematics

1 answer:

Archy [21]3 years ago

8 0

Answer:

Step-by-step explanation:

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sineoko [7]

Answer:

Volume of the tank = 636 ft³

Step-by-step explanation:

Formula for the volume of a cylinder is,

V = πr²h

Where r = radius of the cylinder

h = height of the cylinder

By substituting the values of radius and height of the water tank given in the question,

V = $\pi (4.5)^{2}(10)$

= 202.5π

= 202.5 × (3.14)

= 635.85 ft³

≈ 636 ft³

Therefore, volume of the cylindrical water tank is 636 ft³.

4 0

3 years ago

An open box is to contain a volume of 3 cubic meters. Given that the material forthe sides of the box costs 6 per square meter a

bekas [8.4K]

Answer: total cost = 36/w + 36/l + 9lw

Step-by-step explanation:

The formula for determining the volume of the rectangular box is expressed as

Volume = lwh

Where lwh represents the length, width and height of the box respectively.

Since the volume of the box = 3cm³,then

lwh = 3

h = 3/lw

Since the box is open, the surface area would be

2lh + 2wh + lw

Given that the material for the sides of the box costs 6 per square meter, the total cost for the sides is

(2lh × 6) + (2wh × 6)

Substituting h = 3/lw into the above expression, it becomes.

(2l × 3/lw × 6) + (2w× 3/lw × 6)

= 36/w + 36/l

Since the material for the bottom costs 9per square meter, the total cost for the bottom is

9 × lw = 9lw

Therefore, the total cost of the box as a function of the length land width is

36/w + 36/l + 9lw

7 0

3 years ago

Solve for x: 4 over x equals 5 over 10

julsineya [31]

Answer:

x = 8,

Step-by-step explanation:

$\displaystyle \frac{4}{x} = \frac{5}{10}$ .

$x \ne 0$ for it is a denominator. Multiply both sides with the product of the two denominators: $10\;x$ :

$\displaystyle \frac{4}{x}\cdot (10\; x) = \frac{5}{10}\cdot (10\;x)$ .

$40 = 5\; x$ .

Multiply both sides by one over the coefficient of x:

$\displaystyle \frac{1}{5} \times 40 = \frac{1}{5}\times 5\; x$ .

$x = 8$ .

8 0

4 years ago

Read 2 more answers

Quadrant:

NISA [10]

<h2>✒️Area Between Curves</h2>

$\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}$

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5 0

2 years ago

Zarrin [17]

Answer:

t=-6

Step-by-step explanation:

-4t-4=20 (move constant to the right)

-4t=20+4 (Calculate 20 + 4)

-4t= 24 (then divide 24 on both sides)

5 0

3 years ago