If the ratio of blue marbles to red is 9:4 what is possible number of blue and red marbles in the jar

Which shows the expression (4a ^ 5 * b) ^ 3 simplified?

marshall27 [118]

(4a^5*b)^3 is simplified to
= 64a^15b^3

6 0

3 years ago

Miguel wants to build a container out of sheet metal that has a volume of about 320 cubic inches . He

ZanzabumX [31]

Answer:

cylinder, has the least surface area

Step-by-step explanation:

We are to choose the shape that has the least surface area for the approximate volume desired. In general, the least area for the volume will be provided by a sphere, a "square" cylinder with height equal to diameter, and a cube, in order of increasing area.

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We are asked to find the area and volume of two rectangular prisms, a cylinder, and a square pyramid. Then, we are to identify the shape with the least surface area. Volume and area formulas will be used for the purpose.

<h3>Rectangular Prism</h3>

The relevant formulas are ...

V = LWH

A = 2(LW +H(L +W))

for length L, width W, and height H.

<u>a)</u><u> prism 1</u>

The given dimensions are L = W = 8 in, H = 5 in. Then the volume and area are ...

V = (8 in)(8 in)(5 in) = 320 in³

A = 2((8 in)(8 in) +(5 in)(8 in +8 in)) = 2(64 in² +80 in²) = 288 in²

<u>b)</u><u> prism 2</u>

The given dimensions are L = 10 in, W = 8 in, H = 4 in. Then the volume and area are ...

V = (10 in)(8 in)(4 in) = 320 in³

A = 2((10 in)(8 in) +(4 in)(10 in +8 in)) = 2(80 in² +72 in²) = 304 in²

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<h3>Cylinder</h3>

The relevant formulas are ...

V = πr²h

A = 2πr(r +h)

for radius r and height h.

c) The given dimensions are r = 5 in, h = 4 in. Then the volume and area are ...

V = π(5 in)²(4 in) = 100π in³ ≈ 314 in³

A = 2π(5 in)(5 in +4 in) = 90π in² ≈ 283 in²

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<h3>Square Pyramid</h3>

The relevant formulas are ...

V = 1/3s²h

A = s(s +2H)

for base side dimension s, vertical height h, and slant height H.

d) The given dimensions are s = 10 in, h = 10 in, H = 14 in. Then the volume and area are ...

V = 1/3(10 in)²(10 in) = 1000/3 in³ ≈ 333 in³

A = (10 in)(10 in + 2×14 in) = 380 in²

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<h3>Summary</h3>

The proposed figures have volume and area (rounded to the nearest unit) as follows:

$\begin{tabular}{|c|c|c|c|}\cline{1-4}&shape&V (in^3)&A (in^2)\\\cline{1-4}a&rect prism&320&288\\b&rect prism&320&304\\c&cylinder&314&\bf283\\d&pyramid&333&380\\\cline{1-4}\end{tabular}$

The proposed <em>cylinder</em> requires the least amount of sheet metal for its construction. It has the least surface area of all of the shape choices offered.

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<em>Additional comment</em>

For a volume of 320 in³, a cube would have a surface area of 280.7 in². A "square" cylinder would have an area of 260.0 in². A sphere would have an area of 226.2 in². The above areas are somewhat larger because the shapes depart from the ideal aspect ratio.

3 0

2 years ago

Sarah bought a bike that costs $260. She a coupon that was worth $55 off the cost of any bike. Use the expression $260 + (-55) t

Rudiy27

Answer:

205 dollars for the bike

Step-by-step explanation:

4 0

3 years ago

Evalute 2√72 divided by √8+ √2. I need help asap!!!<br><br><br> :( I will mark brainliest pls!!

Ghella [55]

Answer:

It's 4

Step-by-step explanation:

* / = square root

You simply 2/72 which is 12/2, then simplify /8 which is 2/2.

You combine 2/2 and /2 which is 3/2.

The equation should be 12/2 over 3/2 so you simplify it which will give you 4.

(By simplifying 12/2 over 3/2, you cross out /2 and convert 12 over 3 into a whole number)

6 0

3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago