All categories

4 years ago

Given the volume of Figure A is 512cm ^3and Figure B is 343cm^3, find the ratio of the perimeter from Figure A to Figure B.

1 answer:

5 0

$\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\[2em] \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ \cline{2-4}&\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\[-0.35em] ~\dotfill$

$\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{\textit{figure A}}{\textit{figure B}}\qquad \qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{512}}{\sqrt[3]{343}}\qquad \begin{cases} 512=&2^9\\ &2^{3\cdot 3}\\ &(2^3)^3\\ 343=&7^3 \end{cases}\implies \cfrac{s}{s}=\cfrac{\sqrt[3]{(2^3)^3}}{\sqrt[3]{7^3}} \\\\\\ \cfrac{s}{s}=\cfrac{2^3}{7}\implies \cfrac{s}{s}=\cfrac{8}{7}\implies s:s = 8:7\impliedby \textit{ratio of the }\stackrel{sides~and}{perimeters}$

You might be interested in

Whats 0.07 x 1.22 and explain please with dividing

forsale [732]

ill just multiply it together 0.0854 is the answer with no division

4 0

3 years ago

'y=x+13, xy=30' ????

Art [367]

This is a simultaneous equation. its saying there are two numbers (x and y) that satisfy both equations. Rearrange the 2nd equation to get x=30/y and substitute this into the first equation to get y=(30/y) +13. solve this for y and then you can solve for x.

5 0

3 years ago

Wat is 1000 divided by 350

pochemuha

2.8571 i think im not sure

7 0

3 years ago

Sinker is a weight used by fishermen. A fisherman uses a sinker in the shape of a sphere with radius 0.750.75 cm. The sinker is

omeli [17]

Answer:

Mass m = 20.04 grams

Correct question;

Sinker is a weight used by fishermen. A fisherman uses a sinker in the shape of a sphere with radius 0.75 cm. The sinker is made of lead, which has density 11.34g/cm^3. What is the mass of the sinker?

Step-by-step explanation:

Mass of an object can be defined as

Mass m = density p × Volume V

Given;

Density of the sinker = 11.34g/cm^3

Radius of the sinker = 0.75 cm

And we know that the volume of a sphere is given as;

V = 4/3 × πr^3

Substituting r, we have;

V = 4/3 × π × 0.75^3

V = 1.767 cm^3

Then we can calculate mass as;

m = pV

m = 11.34×1.767

m = 20.03778 g

m = 20.04 grams

6 0

3 years ago

PLEASE HELP ASAP! I don’t recall how to do this!

MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

$(1.05)^x=1.885$

In order to solve for x, we have to get it out from its position of an exponent. Do that by taking the natural log of both sides:

$ln(1.05)^x=ln(1.885)$

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

$x=\frac{ln(1.885)}{ln(1.05)}$

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

$ln(x*x^2)=5$

Simplifying gives you

$ln(x^3)=5$

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

$ln(x)=\frac{5}{3}$

Take the inverse ln by raising each side to e:

$e^{ln(x)}=e^{\frac{5}{3}}$

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

$\frac{1}{2}=e^{.1x}$

"Undo" that e by taking the ln of both sides:

$ln(.5)=ln(e^{.1x})$

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others. Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

$(2^2)^x-6(2)^x=-8$

Now we will bring over the -8 by adding:

$(2^2)^x-6(2)^x+8=0$

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let $u=2^x$

When we do that, we can rewrite the polynomial as

$u^2-6u+8=0$

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor. Now we have to put it back in:

$2^x=4,2^x=2$

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

$2^2=2^x$

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

$2^x=2^1$

Now that the bases are the same, we can say that

x = 1

5 0

3 years ago