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2 years ago

Pls help ASAP i don’t have time!!!!!

Mathematics

1 answer:

Ulleksa [173]2 years ago

6 0

Answer:

i dont have time either >:(

Step-by-step explanation:

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A space shuttle travels at 2.6 x 10,000 feet per second. An hour is 3.6 x 1,000 seconds. This expression can be used to find the

guapka [62]

Answer:

7.22 feet

Step-by-step explanation:

From the question,

s = d/t................. Equation 1

Where s = speed of the space shuttle, d = distance traveled by the shuttle in an hour, t = time taken to travel the distance

make d the subject of the equation,

d = s×t.................... Equation 2

Given: s = 2.6×10000 feet per seconds, t = 1 hour = 3.6×1000 seconds.

Substitute these values into equation 2

d = (2.6×10000)/(3.6×1000)

d = 26000/3600

d = 7.22 feet.

Hence the shuttle travels 7.22 feet in an hour

6 0

2 years ago

Write the relation below as a set of ordered pairs and state if it is a function. If it is not a function, explain why.

AleksAgata [21]

This is a linear function. Let diameter (inches) be x, circumference (inches) be y, observe that the value of y/x is always 3.14. For example, 15.7/5=31.4/10=...=3.14. Therefore, this is not only a function (one to one correspondence from x to y), it's also a linear function that can be represented as y=3.14x.

7 0

2 years ago

Someone help pls -……

uysha [10]

Answer:

d) 35ft

e) 6 seconds

f) 250°

Step-by-step explanation:

If it makes one rotation (360°) in 36 seconds, then it turns 10° every second.

36 secs = 360°

⇒ 1 sec = 360/36 = 10°

d) After 9 seconds, it has turned:

9 x 10 = 90°

Therefore, it will be 15 + 20 = 35ft from the wall

e) 60° = 6 seconds

f) 25 seconds = 25 x 10 = 250°

8 0

1 year 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

Calculeaza:C:23+4x5,17+35,28+5x6,62+2x5,40+10x5.PLLLLZZZZZZZ URGENT

Ann [662]

Answer:the answer is simple it equal 315

Step-by-step explanation:

the first thing you do is to multiply 4x5,5x6,2x5,10x5

7 0

3 years ago