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3 years ago

Find the value of x so the rectangles have the same area

Mathematics

1 answer:

weqwewe [10]3 years ago

7 0

Area of rectangles = width * length.
So x*12 = 12x = area of first rectangle
So 16*(x-2) = 16x -32 = area of second rectangle

The question states to make them equal, so:
first rectangle = second rectangle
12x = 16x -32
Solve for x
-4x = -32
4x = 32
x = 8

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15% of 25 is what? Please answer asap

olganol [36]

Answer: 3.75

Step-by-step explanation:

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2 years ago

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What is an example of when you would want consistent data and, therefore, a small standard deviation?

steposvetlana [31]

Answer:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

$\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271$

And for the standard deviation we can use the following formula:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And after replace we got:

$s = 0.1496$

And as we can ee we got a small value for the deviation <1 on this case.

Step-by-step explanation:

For example if we have the following data:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

We see that the data are similar for all the observations so we would expect a small standard deviation

If we calculate the sample mean we can use the following formula:

$\bar X=\frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271$

And for the standard deviation we can use the following formula:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And after replace we got:

$s = 0.1496$

And as we can ee we got a small value for the deviation <1 on this case.

8 0

2 years ago

The length l, width w, and height h of a box change with time. At a certain instant the dimensions are l = 3 m and w = h = 6 m,

Gemiola [76]

Answer:

a) The rate of change associated with the volume of the box is 54 cubic meters per second, b) The rate of change associated with the surface area of the box is 18 square meters per second, c) The rate of change of the length of the diagonal is -1 meters per second.

Step-by-step explanation:

a) Given that box is a parallelepiped, the volume of the parallelepiped, measured in cubic meters, is represented by this formula:

$V = w \cdot h \cdot l$

Where:

$w$ - Width, measured in meters.

$h$ - Height, measured in meters.

$l$ - Length, measured in meters.

The rate of change in the volume of the box, measured in cubic meters per second, is deducted by deriving the volume function in terms of time:

$\dot V = h\cdot l \cdot \dot w + w\cdot l \cdot \dot h + w\cdot h \cdot \dot l$

Where $\dot w$ , $\dot h$ and $\dot l$ are the rates of change related to the width, height and length, measured in meters per second.

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the volume of the box is:

$\dot V = (6\,m)\cdot (3\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot (3\,m)\cdot \left(-6\,\frac{m}{s} \right)+(6\,m)\cdot (6\,m)\cdot \left(3\,\frac{m}{s}\right)$

$\dot V = 54\,\frac{m^{3}}{s}$

The rate of change associated with the volume of the box is 54 cubic meters per second.

b) The surface area of the parallelepiped, measured in square meters, is represented by this model:

$A_{s} = 2\cdot (w\cdot l + l\cdot h + w\cdot h)$

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time:

$\dot A_{s} = 2\cdot (l+h)\cdot \dot w + 2\cdot (w+h)\cdot \dot l + 2\cdot (w+l)\cdot \dot h$

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the surface area of the box is:

$\dot A_{s} = 2\cdot (6\,m + 3\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m+6\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m + 3\,m)\cdot \left(-6\,\frac{m}{s} \right)$

$\dot A_{s} = 18\,\frac{m^{2}}{s}$

The rate of change associated with the surface area of the box is 18 square meters per second.

c) The length of the diagonal, measured in meters, is represented by the following Pythagorean identity:

$r^{2} = w^{2}+h^{2}+l^{2}$

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time before simplification:

$2\cdot r \cdot \dot r = 2\cdot w \cdot \dot w + 2\cdot h \cdot \dot h + 2\cdot l \cdot \dot l$

$r\cdot \dot r = w\cdot \dot w + h\cdot \dot h + l\cdot \dot l$

$\dot r = \frac{w\cdot \dot w + h \cdot \dot h + l \cdot \dot l}{\sqrt{w^{2}+h^{2}+l^{2}}}$

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the length of the diagonal of the box is:

$\dot r = \frac{(6\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot \left(-6\,\frac{m}{s} \right)+(3\,m)\cdot \left(3\,\frac{m}{s} \right)}{\sqrt{(6\,m)^{2}+(6\,m)^{2}+(3\,m)^{2}}}$

$\dot r = -1\,\frac{m}{s}$

The rate of change of the length of the diagonal is -1 meters per second.

6 0

2 years ago

Line AB passes through points A(–6, 6) and B(12, 3). If the equation of the line is written in slope-intercept form, y = mx + b,

Tresset [83]

(-6,6)(12,3)
slope(m) = (y2 - y1) / (x2 - x1) = (3 - 6) / (12 - (-6) = -3 / 18 = -1/6

y = mx + b
slope(m) = -1/6
use either of ur points (12,3)....x = 12 and y = 3
now sub and find b, the y int
3 = -1/6(12) + b
3 = -12/6 + b
3 = -2 + b
3 + 2 = b
5 = b

so ur slope(m) = -1/6 and ur y int (b) = 5

6 0

3 years ago

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Find the four rational numbers between 2/3 and 5/6

Sav [38]

Answer:

Step-by-step explanation:

LCM of 3 and 6 = 6

2/3 = (2×2) / (3×2) = 4/6

4/6 = (4×5) / (6×5) = 20/30

5/6 = (5×5) / (6×5) = 25/30

Hence, 21/30, 22/30, 23/30 and 24/30 are four rational numbers between 2/3 and 5/6.

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2 years ago