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Brums [2.3K]
3 years ago
7

A contractor is tiling your bathroom with one-foot square tiles. How many tiles are needed for the job? PLS HELPPP

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
4 0

Answer:

Depends on the bathroom

Step-by-step explanation:

Basically since I don't know the bathroom length and width, Ima just cut it down for you.

It is your bathroom area (length*width) divided by the area of the tiles, which is 1*1=1 foot^2(square)

Also it depends on the unit you used for the bathroom, you gotta transfer the unit for the bathroom length and width into feet.

So you should get number for bathroom area with the unit feet square.

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PLEASE HELP WITH GEOMETRY!! WILL GIVE BRAIN!!!
xenn [34]

Answer:

1,726 cm

Step-by-step explanation:

it boils down to this equation:

2(19×29)+2(19×6.4)+2(29×6.5)

bc if you look closely, it's just each side twice (hope that makes sense)

then you work down from there

2(551)+2(123.5)+2(188.5)

1,102+259+337

1,726

5 0
3 years ago
Read 2 more answers
Prove that :-<br><br> Cos3A-cosA/sin3A-sinA+cos2A-cos4A/sin4A-sin2A=sinA/cos2Acos3A
ad-work [718]
\dfrac{\cos^3A-\cos A}{\sin^3A-\sin A}+\dfrac{\cos^2A-\cos^4A}{\sin^4A-\sin^2A}=\dfrac{\sin A}{\cos^2A\cos^3A}\\\\L_s=\dfrac{\cos A(\cos^2A-1)}{\sin A(\sin^2A-1)}+\dfrac{\cos^2A(1-\cos^2A)}{\sin^2A(\sin^2A-1)}\\\\=\dfrac{\cos A(-\sin^2A)}{\sin A(-\cos^2A)}+\dfrac{\cos^2A\sin^2A}{\sin^2A(-\cos^2A)}\\\\=\dfrac{\sin A}{\cos A}-1=\tan A-1

R_s=\dfrac{\sin A}{\cos A\cos^4A}=\dfrac{\sin A}{\cos A}\cdot\dfrac{1}{\cos^4A}=\tan A\cdot\dfrac{1}{\cos^4A}=\dfrac{\tan A}{\cos^4A}


\boxed{L_s\neq R_s}
8 0
3 years ago
Pls Answer this right away! PLS
LenaWriter [7]
The results of the experiment are shown in the frequency side
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3 years ago
Q - 0.63 = 1.16<br> solve for q. please
devlian [24]

Answer: Q=1.79

Step-by-step explanation: Move 0.63 to the right and add both numbers then calculate it. So Q=1.79

7 0
3 years ago
If h (x) = -5x-7 then what is h (x-1) ?
goldenfox [79]

Answer:

h(x - 1) = -5x - 2

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Distributive Property

<u>Algebra I</u>

Terms/Coefficients

Functions

  • Function Notation

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

h(x) = -5x - 7

<u>Step 2: Find</u>

  1. Substitute in <em>x </em>[Function h(x)]:                                                                          h(x - 1) = -5(x - 1) - 7
  2. [Distributive Property] Distribute -5:                                                                h(x - 1) = -5x + 5 - 7
  3. Combine like terms:                                                                                         h(x - 1) = -5x - 2
6 0
3 years ago
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