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2 years ago

7

Jade buys a blouse and a skirt for 34 of their original price. Jade pays a total of $31.50 for the two items. If the original pr

ice of the blouse is $18, what is the original price of the skirt?

1 answer:

frutty [35]2 years ago

3 0

Answer:The original price of the skirt was $16

Step-by-step explanation:18+16=34

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H + 732 = -194 <br> Solve for H

lilavasa [31]

H + 732 = -194

* combine like terms

H = -194 - 732

H = -926

4 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

3 years ago

5. Last year, Dan and Paula regularly invested $25/week in a mutual fund account. This

kozerog [31]

Answer:

They invested total $2730 over the past two years.

Step-by-step explanation:

Consider the provided information.

Dan and Paula regularly invested $25/week in a mutual fund account.

There are 52 weeks in a year.

Total money they invested in a mutual fund in first year.

$25×52 = $1300

They increased their weekly investment by 10 percent.

First find the 10% or 25.

$\$25\times \frac{10}{100}=\$2.5$

That means they invest $25+$2.5 = $27.5

They invest $27.5/week in next year.

Total money they invested in a mutual fund in second year.

$27.5×52 = $1430

Hence the total money they invested in the account over the past two year is:

$1300+$1430=$2730

Hence, they invested total $2730 over the past two years.

8 0

3 years ago

- The lengths (in feet) of the sides of a pentagon can be represented by these

olganol [36]

Answer:

9a+12

Step-by-step explanation:

The perimeter is the sum of the side lengths, so is ...

6a + a + 4 + 8 + 2a

= a(6 +1 +2) + (4 +8)

= 9a +12

8 0

3 years ago

Find the equation of the linear relationship.

rosijanka [135]

Answer:

y = -3x + 150

Step-by-step explanation:

2 points on the graph:

(30, 60) and (40, 30)

Slope:

m=(y2-y1)/(x2-x1)

m=(30 - 60)/(40 - 30)

m= (-30)/10

m= -3

Slope-intercept:

y - y1 = m(x - x1)

y - 60 = -3(x - 30)

y - 60 = -3x + 90

y = -3x + 150

3 0

2 years ago