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3 years ago

Use the cubic model y = 5a3 - 2a2 + a - 45 to find the value of y when x = 4.

1 answer:

3 0

Assuming 5a3 equals 5a^3, and the same for everything else, you would plug in 4 for a, so 5(4)^3 - 2(4)^2 + 4 - 45, which equals 247.

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If ABC FMU, what is the length of MU ?

alina1380 [7]

Fu/ac=mu/bc so 27/36=mu/24 so mu= 18

4 0

3 years ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

3 years ago

22. The area of a triangle can be represented by the expression 14x^5 + 63x^2. If the base is 7x^2, write

kkurt [141]

Answer:

h = 4x ^ 3 + 18

Step-by-step explanation:

The first thing you should know for this case is that the area of a triangle is given by:

A = (1/2) * (b) * (h)

Where,

b: base.

h: height.

Clearing the height we have:

h = ((2) * (A)) / (b)

Substituting the values

h = ((2) * (14x ^ 5 + 63x ^ 2)) / (7x ^ 2)

Simplifying the expression:

h = ((2) * (2x ^ 3 + 9))

h = 4x ^ 3 + 18

answer

an expression to represent its height is

h = 4x ^ 3 + 18

4 0

3 years ago

Help me please hurry :)

Fudgin [204]

Answer:

1. f>30

2. s≥140

Hope This Helps!!!

5 0

2 years ago

Dividing fractions 1/8 / 1/3

agasfer [191]

1/8×3/1=3/8
:)___________

8 0

3 years ago

Read 2 more answers