Ask question

Ask question

All categories

2 years ago

15

The radius of a puddle is claimed to be 12.0 inches, correct to within 0.01 inch. Use linear approximation to estimate the resul

ting error, measured in square inches, in the area of the puddle.

2 answers:

navik [9.2K]2 years ago

8 0

Answer:

I think the answer is 23

Ahat [919]2 years ago

8 0

Answer:

the resulting error is about 0.754 in²

Step-by-step explanation:

A(r) -A(r0) ≈ dA/dr·(r -r0)

The area of a circle is given by ...

A(r) = πr²

so the derivative is

dA/dr = 2πr

and the area error is ...

dA/dr·(r -r0) = 2π(12 in)(0.01 in) = 0.24π in² ≈ 0.754 in²

You might be interested in

What conclusions can be made about the amount of money in each account if f represents Molly's account and g represents her brot

Nat2105 [25]

Answer:

(b) is true

Step-by-step explanation:

Given

Molly

$a = 500$ --- starting balance

$m = 10$ --- monthly rate

Her brother

$a = 100$ ---- starting balance

$r = 10\%$ --- annual rate

Required

Determine which option is true

First, we calculate her brother's function.

The function is an exponential function calculated as:

$y = ab^x$

Where $b = 1 + r$

So, we have:

$y = ab^x$

$y = 100 *(1 + 10\%) ^x$

$y = 100 *(1 + 0.10) ^x$

$y = 100 *(1.10) ^x$

Hence:

$g(x) = 100 *(1.10) ^x$

Next, we calculate Molly's function (a linear function)

The monthly function is:

$y = mx + a$

So, we have:

$y = 10x + 500$

Annually, the function will be:

$y = 10x*12 + 500$

$y = 120x + 500$

So, we have:

$f(x) = 120x + 500$

At this point, we have:

$f(x) = 120x + 500$ ---- Molly

$g(x) = 100 *(1.10) ^x$ ---- Her brother

<u>Next, we test each option</u>

(a): Molly's account will have a faster rate of change over [32,40]

We calculated Molly's function to be:

$y = 120x + 500$

The slope of a linear function with the form: $y = mx + b$ is m

By comparison:

$m = 120$

Since Molly's account is a linear function, the rate of change over any interval will always be the same; i.e.

$m = 120$

For his brother:

Rate of change is calculated using:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(40) - g(32)}{40 - 32}$

$m = \frac{g(40) - g(32)}{8}$

Calculate g(40) and g(32)

$g(x) = 100 *(1.10) ^x$

$g(40) = 100 * 1.10^{40} =4526$

$g(32) = 100 * 1.10^{32} = 2111$

So, we have:

$m = \frac{4526 - 2111}{8}$

$m = \frac{2415}{8}$

$m = 302$

By comparison: $302 > 120$

Hence, her brother's account has a faster rate over [32,40]

(a) is false

(b): Molly's account will have a slower rate of change over [24,30]

$m = 120$ --- Molly's rate of change

For his brother:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(30) - g(24)}{30 - 24}$

$m = \frac{g(30) - g(24)}{6}$

Calculate g(30) and g(24)

$g(x) = 100 *(1.10) ^x$

$g(40) = 100 * 1.10^{30} =1745$

$g(32) = 100 * 1.10^{24} = 985$

So, we have:

$m = \frac{g(30) - g(24)}{6}$

$m = \frac{1745 - 985}{6}$

$m = \frac{760}{6}$

$m = 127$

By comparison: $127 > 120$

Hence, Molly's account has a slower rate over [24,30]

(b) is false

(c): Molly's account will have a slower rate of change over [0,4]

$m = 120$ --- Molly's rate of change

For his brother:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(4) - g(0)}{4 - 0}$

$m = \frac{g(4) - g(0)}{4}$

Calculate g(4) and g(0)

$g(x) = 100 *(1.10) ^x$

$g(4) = 100 * 1.10^4 =146$

$g(0) = 100 * 1.10^{0} = 100$

So, we have:

$m = \frac{g(4) - g(0)}{4}$

$m = \frac{146 - 100}{4}$

$m = \frac{46}{4}$

$m = 11.5$

By comparison: $120>11.5$

Hence, Molly's account has a faster rate over [0,4]

(c) is false

4 0

3 years ago

PLZ HELP ME I NEED HELP

NemiM [27]

Answer:

A,C and B

Step-by-step explanation:

4 0

3 years ago

Compré mi despensa en línea (usando la página de un centro comercial) y en total por los productos que adquirí debía pagar $1600

Mekhanik [1.2K]

Answer:

1.235%

Step-by-step explanation:

Paso 1

Calculamos el costo de envío

= $ 1620 - $ 1600

= $ 20

Paso 2

¿Cuál es el porcentaje de mi compra que me cobraron por los gastos de envío?

Esto se calcula como:

(Costo de envío / Cargos totales) × 100

= 20/1620 × 100

= 1.2345679012%

= Aproximadamente = 1.235%

7 0

2 years ago

Help??? I Don’t Understand

Andrews [41]

The correct answer is line TX and line QX are perpendicular.

5 0

3 years ago

What is 2log5(5x^3)+(1/3)log5((x^2)+6) written as a single logarithm?

joja [24]

Assuming 5 is the base. I'm going to leave that out for now.

2log(5x^3) + (1/3)log(x^2+6)

power rule
log(5^2 x^3*2) + log((x^2 + 6)^(1/3))

log(25x^6) + log((x^2 + 6)^(1/3))

quotient rule
log(25x^6 / (x^2 + 6)^(1/3))

4 0

3 years ago

Read 2 more answers