F(x)=2(0.65)^x is exponential growth or decay

Ayush and sumit take two different routes to school. Ayush's route is 1km 2hm long while sumit route is 2hm 30dam long while rou

emmasim [6.3K]

Answer:

Ayush's route is 0.7 km or 700m longer than Sumit route.

Step-by-step explanation:

Ayush's route is 1km 2hm long while sumit route is 2hm 30dam.

We know that,

1 km = 10 hm

1 dam = 0.1 hm

Using these conversions we get

Ayush's route = 1km 2hm = (1×10) hm + 2 hm = 12 hm

Sumit route = 2hm 30dam = 2 hm + (30×0.1) hm = 2 hm + 3 hm = 5 hm

Ayush's route is longer.

Difference = 12 hm - 5 hm = 7 hm = 0.7 km [1 km = 10 hm]

Hence, Ayush's route is 0.7 km or 700 m longer than Sumit route.

4 0

3 years ago

Read 2 more answers

−4(3m−3)+(−10+8m)=?

Valentin [98]

-4(3m-3)+(-10+8m)
12m+12-1+8m
12m+8m = 20m
+12 -1= +11
answer: 20m+11

*note negative times a negative equals positive and negative times positive is negative
first: you multiply -4 by 3m which is -12m.
second: you multiply -4 by -3 which is +12
third: find the invisible which is 1. (located right here: +(1) (-10+8m)
fourth: multiply invisible 1 by -10 which is -10
fifth: multiply invisible one by +8m which is +8m
sixth: combine like terms
seventh: answer

5 0

3 years ago

If UW = 25 - 35 and UY = S, find the value

Ghella [55]

What is a feeble minded person?

Definition of feebleminded

1 dated, offensive : impaired in intellectual ability : affected with intellectual disability.

Download pdf

5 0

2 years ago

Read 2 more answers

Square lMNO is shown in the diagram below what are the coordinates of the midpoint of a diagonal LN

sashaice [31]

Answer:

The coordinates of the midpoint of LN are $(-2\frac{1}{2},4\frac{1}{2})$

answer (4)

Step-by-step explanation:

* Lets explain how to find the midpoint of a line

- The coordinates of the midpoint of a line whose endpoints are (x1 , y1)

and (x2 , y2) are $x=\frac{x_{1}+x_{2}}{2},y=\frac{y_{1}+y_{2}}{2}$

∵ LMNO is a square

∵ The coordinates of point L are (-6 , 1)

∵ The coordinates of point N are (1 , 8)

- Let the coordinates of point L are (x1 , y1) , the coordinates of point

N are (x2 , y2) and the coordinates of the midpoint of LN are (x , y)

∴ x1 = -6 , x2 = 1 and y1 = 1 , y2 = 8

- Use the rule of the midpoint above to find the midpoint (x , y)

∵ $x=\frac{-6+1}{2}=\frac{-5}{2}=-2\frac{1}{2}$

∵ $y=\frac{1+8}{2}=\frac{9}{4}=4\frac{1}{2}$

∴ The coordinates of the midpoint are $(-2\frac{1}{2},4\frac{1}{2})$

* The coordinates of the midpoint of LN are $(-2\frac{1}{2},4\frac{1}{2})$

8 0

3 years ago

I need help (please show your work)

oksano4ka [1.4K]

Answer:

the solution is (-4, 3)

Step-by-step explanation:

if you were to graph y = -7/4x - 4 you could plot your first point at (0, -4) and get your second point by going up 7 units (from (0, -4) and to the left by 4

if you were to graph y = -1/4x + 2 you could plot your first point at (0, 2) and get your second point by going up 1 unit (from (0, 2) and to the left by 4

the second point plotted for each equation will be the point of intersection, (-4, 3)

6 0

3 years ago