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2 years ago

9

The sides and bottom of the pool need to be tiled. The Tyler always orders 10% more than he needs to allow for breakages. How ma

ny m2 of tiles should the Tiler order

1 answer:

Advocard [28]2 years ago

6 0

Step-by-step explanation:

3 <em><u>poooooooooooooooooooo</u></em>

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Help ASAP! Need full rule

Ulleksa [173]

I dont know this one sorry

3 0

2 years ago

15x+12=45+4x<br> whats x?

saul85 [17]

<span>15x+12=45+4x </span>⇒15x - 4x = 45 - 12 ⇒ 11x = 33 ⇒ x = 3

4 0

3 years ago

Read 2 more answers

A farmer has 520 feet of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one s

Setler [38]

Answer:

$310\text{ feet and }210\text{ feet}$

Step-by-step explanation:

GIVEN: A farmer has $520 \text{ feet}$ of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one side of the pen. The length of the barn is $310 \text{ feet}$ .

TO FIND: Determine the dimensions of the rectangle of maximum area that can be enclosed under these conditions.

SOLUTION:

Let the length of rectangle be $x$ and $y$

perimeter of rectangular pen $=2(x+y)=520\text{ feet}$

$x+y=260$

$y=260-x$

area of rectangular pen $=\text{length}\times\text{width}$

$=xy$

putting value of $y$

$=x(260-x)$

$=260x-x^2$

to maximize $\frac{d \text{(area)}}{dx}=0$

$260-2x=0$

$x=130\text{ feet}$

$y=390\text{ feet}$

but the dimensions must be lesser or equal to than that of barn.

therefore maximum length rectangular pen $=310\text{ feet}$

width of rectangular pen $=210\text{ feet}$

Maximum area of rectangular pen $=310\times210=65100\text{ feet}^2$

Hence maximum area of rectangular pen is $65100\text{ feet}^2$ and dimensions are $310\text{ feet and }210\text{ feet}$

5 0

3 years ago

2. Draw the image of ∆RST under the dilation with scale factor ⅓ and center of dilation at the origin. Label the image ∆R’S’T’.

Alchen [17]

Since center of dilation is the origin, this is easy. Just divide all of the x and y coordinate values by 3. Place the new point on the graph, and draw the triangle.

R' = R(3,6)/3 = (3/3,6/3)=(1,2). So R'(1,2)
S' = S(-3,6)/3 = (-3/3,6/3)=(-1,2). So S'(-1,2)
T' = T(-6,-6)/3 = (-6/3,-6/3)=(-2,-2). So T'(-2,-2)

So you now know the location of the 3 new points. R' at (1,2), S' at (-1,2) and T' at (-2,-2). Simply draw those 3 points on your graph and connect the lines to make a new triangle.

6 0

2 years ago

Two buses leave towns 1631km apart at the same time and travel toward each other. one bus travels 17 /kmh slower than the other.

Blababa [14]

Suppose one bus travels at x km/h, the second bus is 17 km/h slower so it is x-17
the first bus travels a total of 7x km, the second bus travels 7(x-17)km, together they travel the total distance of 1631 km
7x + 7(x-17) =1631
7x + 7x -119 =1631
14x=1750

x=125
x-17=108

5 0

3 years ago