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3 years ago

The answer to my question

Mathematics

2 answers:

AfilCa [17]3 years ago

6 0

C because anything over 1 is the unit rate ex: 4 hour /2 = 2 hour 2 miles /2 = 1 mile

poizon [28]3 years ago

5 0

The answer would be B. because a unit rate is usually equal to one for the sake of conversion

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A car travels at the rate of x miles per hour for y minutes. which of the following gives the number of miles the car has travel

harkovskaia [24]

Given that speed of car = x miles per hour

Time of travel = y minutes = y/60 hours

(Because there are 60 minutes in an hour)

We have to calculate distance travelled so we will use formula

Distance = speed * time

Distance = x * y/60

Distance = xy/60

Hence choice d) xy/60 is the final answer.

7 0

4 years ago

HELP PLEASE DUE IN 3 MINUTES

Morgarella [4.7K]

Answer: it is d becaues that is the mode

4 0

3 years ago

Assuming that the equation defines x and y implicitly as differentiable functions xequalsf(t), yequalsg(t), find the slope of

Doss [256]

Answer:

$\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\$

Hence, the slope , $\dfrac{dy}{dx} = \dfrac{-1}{64}$

Step-by-step explanation:

We need to find the slope, i.e. $\dfrac{dy}{dx}$ .

and all the functions are in terms of $t$ .

So this looks like a job for the 'chain rule', we can write:

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)$

Given the functions

$x = f(t)\\y = g(t)\\$

and

$x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)$

we can differentiate them both w.r.t to $t$

first we'll derivate Eq(B) to find dx/dt

$x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\$

we can also rearrange Eq(B) to find x in terms of t , $x = (37 - 4t^2)^{1/3}$ . This is done so that $\frac{dx}{dt}$ is only in terms of t.

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\\dfrac{dx}{dt} = \dfrac{-8(3)}{3(37 - 4(3)^2)^{2/3}}\\\dfrac{dx}{dt} = -8$

now let's differentiate Eq(C) to find dy/dt

$2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}$

rearrange Eq(C), to find y in terms of t, that is $y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}$ . This is done so that we can replace y in $\frac{dy}{dt}$ to make only in terms of t

$\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}$

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that $\frac{dy}{dt}$ is being multiplied with $\frac{dt}{dx}$ and NOT $\frac{dx}{dt}$ , so we have to use the reciprocal!

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx}\\\dfrac{dy}{dx} = \dfrac{1}{8}.\dfrac{1}{-8} \\\dfrac{dy}{dx} = \dfrac{-1}{64}$

our slope is equal to $\dfrac{-1}{64}$

7 0

3 years ago

Find the median number of magazines purchased in a store by 4 customers. 10, 55 , 34, 21

kkurt [141]

Median of 10, 55, 34, 21

First we arrange number in ascending or descending

10, 21, 34, 55 in ascending.

Median is the middle number and this is between 21 and 34.

We find the average:

(21 + 34) /2 = 55/2 = 27.5

Median is 27.5.

8 0

3 years ago

2x+1-14x=-71 please help me solve this

Alex_Xolod [135]

Step-by-step explanation:

add the like terms

2x - 14x = -71 -1

-12x = -72

12x = 72

x = 6

8 0

4 years ago

Read 2 more answers