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3 years ago

13

3. Melissa and her family went to IHOP for breakfast. There total bill came to $40. The waiter was so good they decided to leave

a 15% tip. What was the amount of the tip

2 answers:

alexandr1967 [171]3 years ago

7 0

15% of 40 is 6 so the answer is $6 I hope this helps!!

lora16 [44]3 years ago

6 0

Answer:

$6

Step-by-step explanation:

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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

In ΔUVW, w = 44 cm, u = 83 cm and ∠V=141°. Find the area of ΔUVW, to the nearest square centimeter.

Tema [17]

Answer:

$Area \approx 1149\ cm^2$

Step-by-step explanation:

<u>Given that:</u>

ΔUVW,

Side w = 44 cm, (It is the side opposite to $\angle W$ )

Side u = 83 cm (It is the side opposite to $\angle U$ )

and ∠V=141°

Please refer to the attached image with labeling of the triangle with the dimensions given.

Area of a triangle with two sides given and angle between the two sides can be formulated as:

$A = \dfrac{1}{2}\times a\times b\times sinC$

Where a and b are the two sides and

$\angle C$ is the angle between the sides a and b

Here we have a = w = 44cm

b = u = 44cm

and ∠C= ∠V=141

Putting the values to find the area:

$A = \dfrac{1}{2}\times 44\times 83\times sin141\\\Rightarrow A = \dfrac{1}{2} \times 3652 \times sin141\\\Rightarrow A =1826 \times 0.629\\\Rightarrow A \approx 1149\ cm^2$

So, the <em>area </em>of given triangle to the nearest square centimetre is:

$Area \approx 1149\ cm^2$

8 0

3 years ago

Kate walks half a mile to the library how many yards does she walk

Crank

Theres 1760 yards in a mile so divide that by 2.
answer is 880

5 0

3 years ago

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What is the area of this figure?

Oksanka [162]

<h3>Answer: 297 square cm</h3>

=============================================

Work Shown:

Break up the figure as shown in the diagram below.

A rectangle and trapezoid form

-------------

Area of rectangle = base*height = 9*15 = 135 cm^2

Area of trapezoid = (1/2)*(b1+b2)*h = (1/2)*(18+9)*12 = 162 cm^2

total area = (area of rectangle)+(area of trapezoid)

total area = (135) + (162)

total area = 297 square cm

3 0

3 years ago

I’m too lazy to do it so if someone can do it it would mean a lot

Dafna11 [192]

Answer:

a

Step-by-step explanation:

did the math :)

brainliest pls!

5 0

3 years ago

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