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2 years ago

Help please important

Mathematics

2 answers:

Kazeer [188]2 years ago

4 0

Answer:

Cos(41) x 30 = Answer

Step-by-step explanation:

Use a calculator to find it out

Lemur [1.5K]2 years ago

3 0

1) tan 41°= 30cm divide by y cm
y= 30 divide by tan-1 41°

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Two out of every 5 costumers at a restaurant use cash to pay for their meal. What percent of the costumers pay with cash?

Gnoma [55]

Answer:

40%

Step-by-step explanation:

Percent means out of 100

Multiply the top and bottom by 20 to get the bottom to 100

2/5 * 20/20 = 40/100

The percent is 40%

6 0

2 years ago

Read 2 more answers

Find the simple interest on a P10500 loan at 20% for two years. (show

Anni [7]

Answer:

P4200

Step-by-step explanation:

SI=PRT/100

P=Principal which is P10500

R=Rate which is 20%

T=Time which is 2 years

So SI= 10500×20×2/100

=P4200

6 0

2 years ago

Someone please help out

exis [7]

Answer:

$39+0.15m \leqslant 50$

Step-by-step explanation:

From the question, Diana wants her total bill to be less than $50.

Diana pays a fixed monthly bill of

$39 for her mobile phone and an additional

$0.15 per minute for phone internet service.

The bill Diana pays for m minutes of using the internet is $0.15m in a month.

The total monthly bill becomes,

$39+0.15m$

The maximum bill she wants to pay is $50.

$\implies39+0.15m \leqslant 50$

This inequality can be solved to determine the number of available minutes.

4 0

2 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

1. Why does the advisor not believe the woman knocking at the town<br> gate is a princess?

Tresset [83]

Answer:

$\pink{\rule{40pt}{555555pt}}$

Step-by-step explanation:

$\pink{\rule{40pt}{555555pt}}$

8 0

2 years ago