All categories

3 years ago

Approximate the square root of 500 to the nearest tenth. Show your work.

1 answer:

4 0

$\sqrt{500} \rightarrow \sqrt{5 * 10 * 10} \rightarrow 10\sqrt{5}$ This is an exact answer. the answer will be a bit larger than 20

You might be interested in

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

What is the constant rate of change in given equation Y= -5x *please Hurry*

statuscvo [17]

Answer:

-5

Step-by-step explanation:

The x and the y are the factors of the equation. Simply extract the value that is not the variable, and, well, you’ll get your constant.

6 0

2 years ago

One plane can travel 30 miles per hour faster than another. one of them goes 185 miles in the same time it takes the other to go

ycow [4]

Plane #1 speed: x mph
Plane #2 speed: (x+30) mph

time = time becomes

170 mi 185 mi
---------- = --------------
x x+30

Solving for x, the speed of the slower plane, we get

170x + 5100 = 185 x
5100
Then 15x = 5100, and x = ---------- (mi/hr) = 340 mph
15

The slower plane flies at 340 mph, and the faster one at 370 mph.

6 0

3 years ago

Please help on number 3

Sedbober [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Find the value of x in each of the following<br>a) 5x/2+ 1 = 11<br>b) 2x/7 – 3 = 2

laila [671]

Answer:

A) x=4

B) x=35/2

Step-by-step explanation:

calculate the product

A) 5x/2+1=11

multiply both sides of the equation by 2

5x+2=22

move the constant to the right hand side and change its sign

5x+22-2

now subtract the numbers

5x=20

now didvide both sides

5x=20

x=4

calculate the product

2x/7-3=2

multiply both sides of the equation by 7

2x-21=14

move the constant to the right hand side and change its sign

2x=14+21

add the numbers

2x=35

now divide both sides of the equation by 2

x=35/2

8 0

3 years ago