All categories

3 years ago

16 centimeters to 44.2 centimeters

Mathematics

1 answer:

Phoenix [80]3 years ago

7 0

Problem work:

y1 =16
y2 = 44.2

( 44.2-16/16) x 100

(1.7625) x 100

So the percent increase (percent change) will be 176.25%

hope this helps

You might be interested in

Tim installs 50 square feet on his floor in 45 minutes. At this rate, how long does it take him to install 495 square feet?

Katena32 [7]

Answer:

445.5 minutes

or 7 hours 25 minutes 30 seconds

Step-by-step explanation:

We can use ratio's to solve

50 ft^2 495 ft^2

-------------- = ------------

45 minutes x minutes

Using cross products

50x = 45*495

50x=22275

Divide each side by 50

50x/50 = 22275/50

x =445.5

445.5 minutes

If we want to convert to hours and minutes, divide by 60

7 hours is 420 minutes

445.5 -420 = 25.5

7 hours 25 minutes

.5 minutes is 30 seconds

7 hours 25 minutes 30 seconds

6 0

2 years ago

PLS CAN SOMEONE HELP ME PLS

EastWind [94]

Answer:

20 m

Step-by-step explanation:

Let the length of the perpendicular dropped from vertex C to segment AB be x

By geometric mean property:

$x = \sqrt{16 \times 9} \\ \\ x = \sqrt{144} \\ \\ x = 12 \: m \\ \\ By \: Pythagoras \: Theorem: \\ \\ Distance \: across \: the \: pond \\ \\ = \sqrt{ {16}^{2} + {12}^{2} } \\ \\ = \sqrt{256 + 144} \\ \\ = \sqrt{400} \\ \\ = 20 \: m$

7 0

2 years ago

Please help with this calculus problem!

ELEN [110]

$\dfrac{\mathrm dQ}{\mathrm dt}=\dfrac{a(1-5bt^2)}{(1+bt^2)^4}$
$Q(t)=\displaystyle\int\frac{a(1-5bt^2)}{(1+bt^2)^4}\,\mathrm dt$

Let $t=\dfrac1{\sqrt b}\tan u$ , so that $\mathrm dt=\dfrac1{\sqrt b}\sec^2u\,\mathrm du$ . Then the integral becomes

$\displaystyle\frac a{\sqrt b}\int\frac{1-5b\left(\frac1{\sqrt b}\tan u\right)^2}{\left(1+b\left(\frac1{\sqrtb}\tan u\right)^2\right)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{(1+\tan^2u)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{\sec^6u}\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(\cos^6u-5\sin^2u\cos^4u)\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(6\cos^2u-5)\cos^4u\,\mathrm du$

One way to proceed from here is to use the power reduction formula for cosine:

$\cos^{2n}x=\left(\dfrac{1+\cos2x}2\right)^n$

You'll end up with

$=\displaystyle\frac a{16\sqrt b}\int(5\cos2u+8\cos4u+3\cos6u)\,\mathrm du$
$=\displaystyle\frac a{16\sqrt b}\left(\frac52\sin2u+2\sin4u+\frac12\sin6u\right)+C$
$=\dfrac a{32\sqrt b}(5\sin2u+4\sin4u+\sin6u)+C$
$Q(t)=\dfrac{at}{(1+bt^2)^3}+C$

6 0

2 years ago

Find a function of the form f(x) = a sin(bx − c) such that the graph of f matches the figure.

aliina [53]

Answer:

y = 3sin(2x+pi)

Step-by-step explanation:

We know that a is equal to 3 because the distance from 3 to the midpoint 0 is 3

We also know that is a phase shift of pi/2 as that is where the new start of our sine function lands

We know there is a compression of 2 as our period has changed from 2pi to pi

So naturally we change pi/2 to pi since we put the 2 inside the parenthesis

Best of luck

4 0

2 years ago

Janelle has d dimes and n nickels. Her sister has 5 times as many times and 6 times as many nickels as Janelle has. Write the su

Molodets [167]

Answer:

Step-by-step explanation:

If Janelle has dimes and nickels, she has d + n. If her sister has 5 times the number of dimes, she has 5d; if her sister has 6 times the number of nickels she has 6n.

d + n + 5d + 6n --> 6d + 7n

7 0

2 years ago

Read 2 more answers