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Elena L [17]
3 years ago
15

Does the size of parachute affect the speed and why

Mathematics
1 answer:
Kryger [21]3 years ago
6 0
Yes it does , more mass area = less space
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10. Given m and b, write the equation of the line:<br><br> m=-4; b=2
steposvetlana [31]

Answer:

y= -4x + 2

Step-by-step explanation:

Y= mx + b is used to find the equation of the line

m=-4 a d b= 2

So, you put them into the equation.

6 0
2 years ago
What is the equation of the following line
Alex_Xolod [135]
Its B 
rise over run dictates that its 2/8 which simplifies to 1/4
and its at an upward angle from left to right therefore its positive.
3 0
3 years ago
Read 2 more answers
Can someone help me pls
qaws [65]

Answer:Example1: 2 and 2 1/2

The mark that is closest to the right end of the glue stick is for 2 1/2 inches.

Answer on last part is 2 1/2 inches

Example 2: 1 2/4 and 1 3/4

The mark that is closest to the right end of the paper clip is for 1 3/4

Answer on last part is 1 3/4

Step-by-step explanation: the ruler is split into halves and fourths between the whole numbers the longer lines are halves shorter are fourths.

5 0
3 years ago
write an equation of the perpendicular bisector of the line segment whose endpoints are (-1,1) and (7,-5)
gladu [14]
Find midpoint of segment.
(-1,1), (7,-5)
x= \frac{-1+7}{2}=3,y= \frac{1+(-5)}{2}=-2

Find slope of line which passes through (-1,1) and (7,-5).
m= \frac{-5-1}{7-(-1)}= \frac{-6}{8}=- \frac{3}{4}

The slope of perpendicular line is negative reciprocal value of m.
m_{\perp}=- \frac{1}{m} =- \frac{1}{- \frac{3}{4} } = \frac{4}{3}

Write quation of line which slope is \frac{4}{3} and passes throug the point (3,-2).
m=\frac{4}{3},x_1=3,y_1=-2&#10;\\y-y_1=m(x-x_1)&#10;\\&#10;\\y-(-2)=\frac{4}{3}(x-3)&#10;\\&#10;\\y+2=\frac{4}{3}x-\frac{8}{3}&#10;\\&#10;\\3y+6=4x-8&#10;\\4x-3y-14=0

3 0
2 years ago
Find the exponential function that passes through the points (2,80) and (5,5120)​
juin [17]

Answer:

  y = 5·4^x

Step-by-step explanation:

If you have two points, (x1, y1) and (x2, y2), whose relationship can be described by the exponential function ...

  y = a·b^x

you can find the values of 'a' and 'b' as follows.

Substitute the given points:

  y1 = a·b^(x1)

  y2 = a·b^(x1)

Divide the second equation by the first:

  y2/y1 = ((ab^(x2))/(ab^(x1)) = b^(x2 -x1)

Take the inverse power (root):

  (y2/y1)^(1/(x2 -x1) = b

Use this value of 'b' to find 'a'. Here, we have solved the first equation for 'a'.

  a = y1/(b^(x1))

In summary:

  • b = (y2/y1)^(1/(x2 -x1))
  • a = y1·b^(-x1)

__

For the problem at hand, (x1, y1) = (2, 80) and (x2, y2) = (5, 5120).

  b = (5120/80)^(1/(5-2)) = ∛64 = 4

  a = 80·4^(-2) = 80/16 = 5

The exponential function is ...

  y = 5·4^x

3 0
2 years ago
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