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3 years ago

Jen picked three fourth in half an hour if she keeps the same rate how many will she pick in 2 hours

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

5 0

Alright, lets get started.

Jen picked three fourth in half an hour.

Half an hour means 30 minutes.

At the same rate, how many she can pick in 2 hours.

2 hours means $2 * 60 = 120$ minutes

Now,

In 30minutes, Jen can pick = $\frac{3}{4}$

In 1 minute, Jen can pick = $\frac{3}{4} *\frac{1}{30}$

In 120 minutes, Jen can pick = $\frac{3}{4} *\frac{1}{30} * 120 = 3$

Hence Jen can pick complete 3 in 2 hours of time.

So the answer is 3 : Answer

Hope it will help :)

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Given the second order homogeneous constant coefficient equation y′′−4y′−12y=0 1) the characteristic polynomial ar2+br+c is r^2-

Vitek1552 [10]

Answer:

The initial value problem $y(x) = 4 e^{-2x} -3 e^{6 x}$

Step-by-step explanation:

Step1:-

a) Given second order homogenous constant co-efficient equation

$y^{ll} - 4y^{l}-12y=0$

Given equation in the operator form is $(D^{2} -4D-12)y=0$

Step 2:-

b) Let f(D) = $(D^{2} -4D-12)y$

Then the auxiliary equation is $(m^{2} -4m-12)=0$

Find the factors of the auxiliary equation is

$m^{2} -6m+2m-12=0$

m(m-6) + 2(m-6) =0

m+2 =0 and m-6=0

m=-2 and m=6

The roots are real and different

The general solution $y = c_{1} e^{-m_{1} x} + c_{2} e^{m_{2} x}$

the roots are $m_{1} = -2 and m_{2} = 6$

The general solution of given differential equation is

$y = c_{1} e^{-2x} + c_{2} e^{6 x}$

Step 3:-

C) Given initial conditions are y(0) =1 and y1 (0) =-26

The general solution of given differential equation is

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$ .....(1)

substitute x =0 and y(0) =1

$y(0) = c_{1} e^{0} + c_{2} e^{0}$

$1 = c_{1} + c_{2}$ .........(2)

Differentiating equation (1) with respective to 'x'

$y^l(x) = -2c_{1} e^{-2x} + 6c_{2} e^{6 x}$

substitute x= o and y1 (0) =-26

$-26 = -2c_{1} e^{0} + 6c_{2} e^{0}$

$-2c_{1} + 6c_{2} = -26$ .............(3)

solving (2) and (3) by using substitution method

$substitute c_{2} =1- c_{1}$ in equation (3)

$-2c_{1} + 6(1-c_{1}) = -26$

on simplification , we get

$-2c_{1} + 6(1)-6c_{1}) = -26$

$-8c_{1} = -32$

dividing by'8' we get $c_{1} =4$

substitute $c_{1} =4$ in equation $1 = c_{1} + c_{2}$

so $c_{2} = 1-4 = -3$

now substitute $c_{1} =4 and c_{2} =-3$ in general solution

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$

$y(x) = 4 e^{-2x} -3 e^{6 x}$

now the initial value problem

$y(x) = 4 e^{-2x} -3 e^{6 x}$

7 0

3 years ago

The graph shows the relationship between the total cost and the number of gift cards that Raj bought for raffle prizes.

BabaBlast [244]

The problem is as following:

The graph shows the relationship between the total cost and the number of gift cards that Raj bought for raffle prizes. What would be the cost for 5 of the gift cards? $80, $90, $100 Or $110

===============================================================

Solution:

The points of the figure represents a straight line

The general form of the straight line is ⇒ y = mx + c

where m is the slope and c is constant

let x ⇒ The number of gifts cards , y ⇒ Total cost (in $)

we will find m and c using the points (1,20) , (2,40)

The slope = m = $\frac{y_{2} - y_{1}}{x_{2}-x_{1}} =\frac{40-20}{2-1} =\frac{20}{1}=20$

∴ y = 20x + c

Substitute with the point (1,20) ⇒ y = 20 at x = 1

∴ 20 = 20 * 1 + c

∴ 20 = 20 + c

∴ c = 0

∴ The equation of the line ⇒⇒⇒ y = 20x

To find the cost for 5 of the gift cards substitute with x = 5

∴ y = 20 * 5 = 100

So, the cost for 5 of the gift cards = $100

The correct answer is option 3 ⇒⇒⇒ $100

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Answer:

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Answer:

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Answer:

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Step-by-step explanation:

Since the cost for border paper is given in terms of feet and the dimensions of the gym are given in meters, we want to convert them to the same unit. Let's choose to convert feet to meters since that would be only one conversion:

$15\:\mathrm{ft}=4.572\:\mathrm{m}$

Therefore, the cost of border paper is $6.99 pet 15 feet or $6.99 pet 4.572 meters.

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$p=15+27+15+27,\\p=2(15)+2(27)=84\:\mathrm{m}$

Therefore, the cost of the total border paper for the gym is equal to:

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