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2 years ago

Find each unit rate. 1. 3/8 mile in 3/4 hour.

Mathematics

1 answer:

MaRussiya [10]2 years ago

6 0

3/8 : 3/4
=(3/8)(4/3)
=1/2 mile per hour
=0.5 mile per hour

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The length of a rectangle is 12 units longer than the width. The perimeter is 7 times the width. Find the length and the width o

katen-ka-za [31]

Log explanation below; answer is at bottom.

If the length, l, is 12 units longer than the width, w, then w = l - 12. If the perimeter, p,

p = 2l +2w,

is 7 x w, then

w = p/7

When two things are set equal to the same variable, they are equal to each other, so,

l - 12 = p/7

Now you need to get rid of the p so you are only working with one variable. To do this you plug in whatever p is equal to for p, so,

l - 12 = (2l + 2w)/7 now to get rid of the w do the same thing we did with p just for w. So,

l - 12 = (2l + 2(l - 12))/7

To solve this you want to multiply both sides by 7 first to get rid of the fraction.

7l - 84 = 2l + 2(l - 12)

Next you want to distribute the 2 over the l and the 12.

7l - 84 = 2l + 2l - 24

Next you want to combine like terms on each side.

7l - 84 = 4l - 24

Next add 84 and subtract 4l from sides to isolate the variable.

3l = 60

Now divide each side by 11 to get your answer.

l = 20.

To find the width,

w = l - 12

Just plug in and solve.

w = 20 - 12

w = 8

So your length and width are 20 and 8.

l = 20

w = 8

7 0

3 years ago

What’s answer 1 and 2?

emmasim [6.3K]

hey sorry I need a free point !

4 0

2 years ago

Complete the Point-Slope equation of (-8,-8) (-7,9)

Vadim26 [7]

Answer:

y = 17x +128

Step-by-step explanation:

P1(-8,-8)

P2(-7,9)

Slope,

m = (y2-y1)/(x2-x1) = (9-(-8))/(-8-(-7)) = 17/(-1) = 17

The point-slope equation is

y-y1 = m(x+x1)

y-(-8) = 17(x-(-8))

y+8 = 17x + 136

y = 17x +128

7 0

3 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

Find the product. 2 1/3 . 2 2/3 ? please show your work ty . A) 6 2/9 B) 5 C ) 1 2/3 D) 4 2/9

koban [17]

The answer is 6.2(A) Turn the mixed numbers into improper fractions b multiplying the whole number by the denominator. Do this for both fractions then multiply normally.<span />

6 0

3 years ago