Ask question

Ask question

All categories

2 years ago

5

A builder was building a fence. In the moring he worked 2/5 of an hour. In the afternoon he wirked for 9/10 of an hour how many

times as long in the morning did he work in the afternoon

1 answer:

Burka [1]2 years ago

4 0

Answer:

2 1/4 times

Step-by-step explanation:

A builder was building a fence. In the moring he worked 2/5 of an hour. In the afternoon he wirked for 9/10 of an hour. how many times as long in the morning did he work in the afternoon

Note that:

1 hour = 60 minutes

In the moring he worked 2/5 of an hour.

= 2/5 × 60 minutes

= 24 minutes

In the afternoon he worked for 9/10 of an hour

Hence:

9/10 × 60 minutes

= 54 minutes

How many times as long in the morning did he work in the afternoon?

This is calculated as number of minutes worked in the afternoon ÷ Number of minutes worked in the morning

= 54 minutes/24 minutes

= 2 6/24

= 2 1/4 times

You might be interested in

Jane moved from a house with 72 square feet of closet space to an apartment with 44.64 square feet of closet space. What is the

Rashid [163]

Answer:

c. 38$

Step-by-step explanation:

6 0

2 years ago

Please explain and show work

Vinvika [58]

1*3/5 = 3/5.
3/5(t-6) = -0.4
Distribute 3/5
3/5t - 3.6 = -0.4
Add 3.6
3/5t = 3.2
Divide by 3/5
t = 5 1/3 or 5.33

6 0

3 years ago

Read 2 more answers

Rebecca has 8 basketball cards. This is 1/5 the amount that Eden has. How many basketball cards does Eden have?

Margarita [4]

Answer:

Eden has 40 basketball cards

Step-by-step explanation:

Just do 5*8 which equals to 40, you can go back and check your answer by finding the 1/5 of 40 and you would receive 8

7 0

3 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

2 years ago

Dados os vectores ⃗⃗ = (2, −4), ⃗ = (−5, 1) e ⃗⃗ = (−12, 6), determinar 1 e 2 tal que

Ostrovityanka [42]

sus sus sus sus sus sus sus sus sus sus sus sus sus sus sus sus

3 0

2 years ago