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3 years ago

Please HELPPP!!! Pleaseeeee

Mathematics

1 answer:

Sloan [31]3 years ago

8 0

Step-by-step explanation:

isosceles triangle (|>) this but right side up

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This graph looks like a trigonometric function. Which trigonometric function would be best to use for this model: cosine, sine,

Trava [24]

Sine wave with amplitude 0.25
Period is 4/3
h(t) = 0.25Sin((2Pi/4/3)t) = 0.25Sin((3Pi/2)t)

5 0

3 years ago

Please help!! I'll give 30 points and brainliest.

Valentin [98]

Step-by-step explanation:

$3 + \frac{9}{4} = \frac{9}{x+4} \\ \\ \therefore \: \frac{3 \times 4 + 9}{4} = \frac{9}{x+4} \\ \\ \therefore \: \frac{12 + 9}{4} = \frac{9}{x+4} \\ \\ \therefore \: \frac{21}{4} = \frac{9}{x+4} \\ \\ \therefore \:21(x + 4) = 36 \\ \\ \therefore \:21x + 84 = 36\\ \\ \therefore \:21x = 36 - 84\\ \\ \therefore \:21x = - 48\\ \\ \therefore \:x = \frac{-48}{21} \\ \\ \therefore \:x = \frac{-16}{7} \\ \\ \huge \red{ \boxed{ \therefore \:x = - 2 \frac{2}{7} }}\\ \\$

8 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:

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