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2 years ago

Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Anon25 [30]2 years ago

3 0

Answer:

A) 40

B) 20

Step-by-step explanation:

By <em>Length · Width · Height</em>

A) (5) · (4) · (2) = 40

For part B, the length and width are the same, but the depth (height) of the water is only one foot, so we can replace the height value from the first equation with 1.

B) (5) · (4) · (1) = 20

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I have found the area but I'm not sure how to find the perimeter.

Lana71 [14]

Answer:

20+20+24+24=88

88/2=44

88+44=132

Step-by-step explanation:

If you split it so there is a rectangle and a triangle, you find the perimeter of the rectangle then divide it in half.

6 0

3 years ago

I need help with this?

Dmitrij [34]

el triangulo que forma el radio es un triangulo equilátero y por lo tanto el lado del hexágono sera de la misma medida que el radio:

lado = radio = 2 cm

como hay 6 lados, lo multiplicamos por 6 :

2x6 = 12 cm

entonces el perímetro del hexágono es 12 cm

Saludos¡

6 0

2 years ago

Read 2 more answers

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

Which pairs of rectangles are similar polygons?<br> Select each correct answer.

iragen [17]

Answer:

The bottom two

Step-by-step explanation:

[] Similar beings their sides have equivalent ratios to each other

[] See attached

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

8 0

2 years ago

Hamburgers come packed 12 in a package. Hamburger buns come packed 10 in a package. If we want one hamburger for each bun for a

BlackZzzverrR [31]

Answer:

5 hamburger packages and 6 bun packages

Step-by-step explanation:

It is asking for the highest common multiple of 12 and 10. the first time a multiple of 12 is divisible by 10 is at 12x5, which is 60. now divide 60 by 12 and 10, 5 and 6.

4 0

1 year ago