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3 years ago

10

Kimberly and Johannesburg are 460km/hour apart. A train covers the distance in 5 hours and 45 minutes. Determine the speed of th

e train

1 answer:

ladessa [460]3 years ago

6 0

It traveled 84.404 km/h

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Find the value of x. PLEASE explain how you got your answer thanks!

Mnenie [13.5K]

Answer:

x = 96

Step-by-step explanation:

x + x - 5 + x - 11 + x - 8 = 360

4x - 24 = 360

4x = 360 + 24

4x = 384

x = 96

8 0

3 years ago

I need help with this

11111nata11111 [884]

Answer:

151

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What's the volume and surface area of these two cylinders?

Vesnalui [34]

I swear im not trying to just get points but is this like a pick which is correct cause my answer is <span>Cylinder B has a diameter of 6 inches and a height of 4 inches.</span>

4 0

4 years ago

To buy a car, you borrow $25,000 with a term of three years at an APR of 6.5%. What is your monthly

Finger [1]

Answer:

$766.23

Step-by-step explanation:

The amortization formula is appropriate for this. (Better, use a financial calculator or spreadsheet.)

A = P(r/12)/(1 -(1 +r/12)^-(12t))

where P is the amount borrowed (25,000), r is the annual rate (0.065), and t is the number of years (3).

Putting the given numbers into the formula, we have ...

A = $25,000(0.065/12)/(1 -(1 +0.065/12)^(-36))

A = $25,000(.00541667)/(1 -1.00541667^-36)

A = $25,000(0.00541667)/(0.1767322)

A = $766.23

The monthly payment is $766.23.

_____

<em>Comment on tools</em>

A graphing or scientific calculator is useful for these. Something like a TI-84, or any of the equivalents, will have financial calculations like this built in. There are also numerous apps available for phone or tablet that will do financial calculations.

4 0

3 years ago

For fun question

ZanzabumX [31]

Consider the function $f(x)=x^{1/3}$ , which has derivative $f'(x)=\dfrac13x^{-2/3}$ .

The linear approximation of $f(x)$ for some value $x$ within a neighborhood of $x=c$ is given by

$f(x)\approx f'(c)(x-c)+f(c)$

Let $c=64$ . Then $(63.97)^{1/3}$ can be estimated to be

$f(63.97)\approxf'(64)(63.97-64)+f(64)$
$\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375$

Since $f'(x)>0$ for $x>0$ , it follows that $f(x)$ must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function $f(x)$ . This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...

4 0

3 years ago