In triangle LMN, P lies on LM and Q lies on LN. If Pq is 1/2Mn, PQ is called what

A brand of clay brick tiles has the dimensions 3 5 8 by 7 5 8 by 1 2 inch. These thin bricks are commonly used for floors. Estim

raketka [301]

Answer: The cost to tile the floor is $318.24.

Step-by-step explanation:

Since we have given that

Dimensions of floor is 12 by 17 foot.

So, Area of floor would be

$12\times 17\\\\=204\ sq.\ foot$

As we know that 1 foot = 12 inch

So, Area of floor would be

$204\times (12)^2\\\\=204\times 144\\\\=29376\ sq.\ foot$

Dimension of brick tiles is given by

$3\dfrac{5}{8}\ by\ 7\dfrac{5}{8}\ by\ \dfrac{1}{2}$

So, Area of brick tiles is given by

$2(lb+bh+lh)\\\\=2(\dfrac{29}{8}\times \dfrac{61}{8}+\dfrac{61}{8}\times \dfrac{1}{2}+\dfrac{29}{8}\times \dfrac{1}{2})\\\\=2(\dfrac{1769}{64}+\dfrac{61}{16}+\dfrac{29}{16})\\\\=2\times \dfrac{1769+244+116}{64}\\\\=\dfrac{2129}{32}$

So, Number of clay brick would be

$29376\div \dfrac{2129}{32}\\\\=29376\times \dfrac{32}{2129}\\\\=\dfrac{29376\times 32}{2129}\\\\=441.53\\\\\approx 442$

cost of each clay brick = $0.72

So, Total cost of flooring would be

$442\times 0.72\\\\=\$318.24$

Hence, The cost to tile the floor is $318.24.

7 0

3 years ago

Determine the volume in litres of 14kg of petrol of density 700kg/m³

valkas [14]

Answer:

$p = \frac{m}{v}$

Step-by-step explanation:

Density is the ratio of an object's mass to its volume, and it measures how compact an object is. The density of an object, p, is directly proportional to its mass, m, and inversely proportional to its volume, V:

$p = \frac{m}{v}$

4 0

2 years ago

Please help me no one even bothers to help so plzzzz i BEGGGGGGGGGGG

user100 [1]

Answer:

Here's what I get.

Step-by-step explanation:

When you dilate an object by a scale factor, you multiply its coordinates by the same number.

If the scale factor is 4, the rule is (x, y) ⟶ (4x, 4y)

Your table then becomes

$\begin{array}{lcl}\textbf{Vertices of}& \, & \textbf{Vertices of}\\\textbf{VWXY}& \, & \textbf{V'W'X'Y'}\\V(-1 ,1) & \quad & V'(-4, 4)\\W(-1, 2) & \quad & W'(-4, 8)\\X(2, -1) & \quad & X'(8 ,-4)\\Y(2, 2) & \quad & Y'(8, 8)\\\end{array}$

The diagram below shows figure VVXY as a green bow-tie and its image V'W'X'Y' in orange.

The scale factor is greater than one, so the dilation is an enlargement.

7 0

3 years ago

A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig

Vaselesa [24]

Answer:

a) Depth changing rate of change is $0.24m/min$ , When the water is 6 meters deep

b) The width of the top of the water is changing at a rate of $0.17m/min$ , When the water is 6 meters deep

Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them $\frac{3.5}{10} =\frac{a}{h}$ , then $a=0.35h$ .

a) As we have that volume is $V=\frac{1}{2} 2ahL=ahL$ , then $V=(0.35h^{2})L$ , so we can derivate it $\frac{dV}{dt}=2(0.35h)L\frac{dh}{dt}$ due to the chain rule, then we clean this expression for $\frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt}$ and compute with the knowns $\frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min$ , is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is $2a=0.7h$ , we can derivate it and obtain $\frac{da}{dt}=0.7\frac{dh}{dt} =0.7*0.24m/min=0.17m/min$ The width of the top of the water is changing, When the water is 6 meters deep at this rate

8 0

3 years ago

25 POINT MATH QUESTION HELP ASAP

Ivanshal [37]

A.32 it’s not very complicated you just have to find the value of A and B then multiply them together

8 0

3 years ago