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3 years ago

Which piecewise function is shown on the graph?

Mathematics

2 answers:

Scilla [17]3 years ago

9 0

The very first one probably

harkovskaia [24]3 years ago

7 0

Answer:A

Step-by-step explanation:

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Can somebody help me please ?? :) Thank u if u do

Lorico [155]

SF = 9/18 = 1/2 or 0.5

answer
1/2 or 0.5

5 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

Bamboo Plant A is 1.2 meters tall and growing at a rate of 0.45 meter per day. Bamboo Plant B is 0.85 meter tall and growing 0.5

pashok25 [27]

Answer:

7 days

Step-by-step explanation:

Bamboo plant A = 1.2 + 0.45d

Bamboo plant B = 0.85 + 0.5d

Where,

d = number of days

Equate the two growth

Bamboo plant A = Bamboo plant B

1.2 + 0.45d = 0.85 + 0.5d

Collect like terms

1.2 - 0.85 = 0.5d - 0.45d

0.35 = 0.05d

Divide both sides by 0.05

d = 0.35 / 0.05

d = 7 days

Plant B be taller than Plant A after 7 days

3 0

2 years ago

Kelly buys a sweater for $16.79 and a pair of pants for $28.49. She pays with a $50 bill. How much change should Kelly get back?

andrew11 [14]

16.79+28.49=45.28

50-45.28=4.72

Kelly would get $4.75 back

4 0

3 years ago

Y=-x+3 y = -2.x +7 = 5.

lyudmila [28]

Answer:

Slope: 3/5

y-intercept: 7

Step-by-step explanation:

5 0

3 years ago