All categories

2 years ago

How long would it take to lay 8 rows of 18 bricks each at a rate of 4 bricks per minute

2 answers:

8 0

8 times 18 is 144. then take that and divide it by 60 seconds which equals a minute. your answer is 2 minutes and 40 seconds

erastovalidia [21]2 years ago

3 0

Do 8 times 18 and that equals 144 and then divide that by 4 and that is 36 so your answer is 36

You might be interested in

What is the slope between the following problems? (8,-4) and (2,-4)

Varvara68 [4.7K]

Answer:

Slope = 0/-6

(Sorry for the bad handwriting)

7 0

2 years ago

Read 2 more answers

Guys I need the answer ASAP Is it A.4 B.5 C.6 D.7

Anna71 [15]

Answer:

Step-by-step explanation:

srry if im wrong

7 0

2 years ago

Read 2 more answers

Radioactive Decay:

Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

Answer: The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

Step-by-step explanation:

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}$

The formula used to calculate the time period for a first order reaction follows:

$t=\frac{2.303}{k}\log \frac{a}{(a-x)}$

where,

k = rate constant = $0.01506days^{-1}$

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

$t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days$

Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.

7 0

2 years ago

a circular oil slick is expanding at a rate of 2m^2/h. Find the rate at which it's diameter is expanding when it's radius is 1.5

aleksandrvk [35]

Answer: $\frac{4\pi}{3} \text{ meter per hour}$

Step-by-step explanation:

The circular oil slick is expanding at a rate of $2 m^2/h$

Let A be the area of the circular oil slick,

So, the changes in A with respect to time (t),

$\frac{dA}{dt} = 2$

$\frac{d(\pi r^2)}{dt} = 2$

$2\pi r\frac{dr}{dt} = 2$

$\frac{dr}{dt} = \frac{1}{\pi r}$

Also, the change in diameter with respect to time(t),

$\frac{d}{dt} (2 r) = 2 \frac{dr}{dt}$

$\frac{d}{dt} (2 r) = 2 \times \frac{1}{\pi r}$

$\frac{d}{dt} (2 r) = \frac{2}{\pi r}$

For r = 1.5 m,

$\frac{d}{dt} ( 2 r)]_{r=1.5} = \frac{2}{\pi \times 1.5}=\frac{20}{\pi \times 15}=\frac{4\pi }{3}\text{ meter per hour}$

7 0

2 years ago

Find the domain of the function. y = <img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20%2B%201" id="TexFormula1" title

Step2247 [10]

Answer:

[0, infinity)

Step-by-step explanation:

A square root can only use positive numbers, otherwise it becomes imaginary. A function needs to be real to output real values. Therefore, the domain is everything from 0 onwards.

5 0

3 years ago