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Regina, courtney, and Ellen hiked around bear pond. Regina hiked 7/10 of the distance in an hour. Ellen hiked 3/8 of the distanc

Kitty [74]

7/10 > 3/8

I found this by first finding a common denominator, so for this it will be 40. 10 can go into 40 four times, so you do 7x4 and that is 28, so 7/10 is 28/40.
3/8 is 15/40 because 8 goes into 40 five times, so you do 5x3 which is 15, so the fraction is 15/40. 15/40 is less than 28/40 because 28 > 15

8 0

3 years ago

Sm1 please help me w this question!!!

Lady_Fox [76]

Answer:

The answer is A, (1,2).

Step-by-step explanation:

Hope this helps!

8x-4(x+1) = 0

x =1

y = 1 + 1

y = 2

(x,y) = (1,2)

2 = 1 + 1

8 x 1 - 4 x 2 = 0

Simplify

2 = 2

0 = 0

Solution = (1 , 2)

6 0

3 years ago

01:57:09 In a city, the distance between the library and the police station is 3 miles less than twice the distance between the

tester [92]

Answer:

Distance between the police station and the fire station be 4 miles

Step-by-step explanation:

Let distance between the police station and the fire station be x miles.

The distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station.

Distance between the library and the police station $=2x-3$

Also, distance between the library and the police station is 5 miles.

$2x-3=5\\2x=3+5\\2x=8\\x=\frac{8}{2}\\ x=4\,\,miles$

8 0

3 years ago

5/9 divided by 1/4 in simplest form

natita [175]

20/9 or 2 2/9 5/9 divided by 1/4 = 5/9 x 4 = 20/9

5 0

3 years ago

Read 2 more answers

The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

7 0

3 years ago