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Sliva [168]
3 years ago
7

A pool measuring 8 meters by 26 meters is surrounded by a path of uniform​ width, as shown in the figure. If the area of the poo

l and the path combined is 1008 square​ meters, what is the width of the​ path?
Mathematics
1 answer:
Ainat [17]3 years ago
8 0
Let the width path be x.
Length of the outer rectangle = 26 + 2x.
Width of the outer rectangle = 8 +2x.

Combined Area = (2x + 26)*(2x + 8) = 1008

2x*(2x + 8) + 26*(2x + 8 ) = 1008

4x² + 16x + 52x + 208 = 1008

4x² + 68x + 208 - 1008 = 0
4x² + 68x - 800 = 0.          Divide through by 4.
x²  + 17x - 200 = 0 . This is a quadratic equation.

Multiply first and last coefficients:  1*-200 = -200

We look for two numbers that multiply to give -200, and add to give +17

Those two numbers are 25 and -8.

Check:   25*-8 = -200         25 + -8 = 17

We replace the middle term of +17x in the quadratic expression with 25x -8x


 x² +17x - 200 = 0     

x² + 25x - 8x - 200 = 0     

x(x + 25) - 8(x + 25) = 0

(x+25)(x -8) = 0

x + 25 = 0    or   x - 8 = 0

x = 0 -25              x = 0 + 8

x = -25                    x = 8

The width of the path can not be negative.

The only valid solution is x = 8.

The width of the path is 8 meters.
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What is the surface area of the net of the rectangular prism?
swat32

Answer:

638.32 yd²

Step-by-step explanation:

The surface area of any figure can be found by calculating the area of all the sides and finding the sum.  In a rectangular pyramid, there is a rectangular base (l x w) and the same two triangles on one side and another two identical triangles on the other side.

Area of rectangle: A = l x w

Area of a triangle:  A = 1/2(b)(h)

Rectangle: 16 x 13.4 = 214.4 yd²

Bottom/top triangles: 14.1 x 16 = 225.6 yd²

Side triangles: 14.8 x 13.4 = 198.32 yd²

Total:  198.32 + 225.6 + 214.4 = 638.32 yd²

8 0
3 years ago
Find the tenth term in each sequence. -1,2/3,7/3,4,17/3
Luba_88 [7]

It'a an arithmetic sequence:

a_1=-1\\\\a_2=-1+\dfrac{5}{3}=-\dfrac{3}{3}+\dfrac{5}{3}=\dfrac{2}{3}\\\\a_3=\dfrac{2}{3}+\dfrac{5}{3}=\dfrac{7}{3}\\\\a_4=\dfrac{7}{3}+\dfrac{5}{3}=\dfrac{12}{3}=4\\\\a_5=4+\dfrac{5}{3}=\dfrac{12}{3}+\dfrac{5}{3}=\dfrac{17}{3}\\\vdots\\\\\text{The general formula for nth term of an arithmetic sequence}\\\\a_n=a_1+(n-1)d\\\\\text{We have}\ a_1=-1\ \text{and}\ d=\dfrac{5}{3}.\\\\\text{Substitute}\\\\a_n=-1+(n-1)\left(\dfrac{5}{3}\right)=-1+\dfrac{5}{3}n-\dfrac{5}{3}=\dfrac{5}{3}n-\dfrac{3}{3}-\dfrac{5}{3}

\boxed{a_n=\dfrac{5}{3}n-\dfrac{8}{3}}

5 0
3 years ago
Consider the graph of the quadratic function y = 3x2 – 3x – 6. What are the solutions of the quadratic equation 0 = 3x2 − 3x − 6
son4ous [18]
Y = 3x^2 - 3x - 6 {the x^2 (x squared) makes it a quadratic formula, and I'm assuming this is what you meant...}

This is derived from:
y = ax^2 + bx + c

So, by using the 'sum and product' rule:

a × c = 3 × (-6) = -18

b = -3

Now, we find the 'sum' and the 'product' of these two numbers, where b is the 'sum' and a × c is the 'product':

The two numbers are: -6 and 3

Proof:

-6 × 3 = -18 {product}

-6 + 3 = -3 {sum}

Now, since a > 1, we divide a from the results

-6/a = -6/3 = -2

3/a = 3/3 = 1

We then implement these numbers into our equation:

(x - 2) × (x + 1) = 0 {derived from 3x^2 - 3x - 6 = 0}

To find x, we make x the subject of 0:

x - 2 = 0

OR

x + 1 = 0

Therefore:

x = 2

OR

x = -1

So the x-intercepts of the quadratic formula (or solutions to equation 3x^2 - 3x -6 = 0, to put it into your words) are 2 and -1.


We can check this by substituting the values for x:

Let's start with x = 2:

y = 3(2)^2 - 3(2) - 6
= 3(4) - 6 - 6
= 12 - 6 - 6
= 0 {so when x = 2, y = 0, which is correct}

For when x = -1:

y = 3(-1)^2 - 3(-1) - 6
= 3(1) + 3 - 6
= 3 + 3 - 6
= 0 {so when x = -1, y = 0, which is correct}
7 0
3 years ago
Read 2 more answers
Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS. P(1, 0, −1), Q(3, 3, 0), R(3, −3, 0), S(1, −2, 2)
topjm [15]

Answer:

the volume of the parallelepiped is = 36  

Step-by-step explanation:

given,

P(1, 0, −1),      Q(3, 3, 0),      R(3, −3, 0),     S(1, −2, 2)

PQ = Q - P = (2, 3, 1)

PR = R - P  = (2, -3 , 1)

PS = S - P  = (0, -2 , 3)

now volume of parallelopiped

[PQ PR PS] = \begin{bmatrix}2 & 3 &1 \\ 2 & -3 &1 \\ 0 & -2 &3 \end{bmatrix}

now calculating determinant of the matrix

        = 2 (-9+2) -  3 (6-0) + 1 (-4-0)

        = -14  - 18  - 4

        = -36

hence , the volume of the parallelepiped is = 36  

4 0
3 years ago
I need help plz on this
Amanda [17]
The answer is B.
6y=-4x+8
y=(-4x+8)/6
y=-2/3x+4/3
7 0
3 years ago
Read 2 more answers
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