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2 years ago

Aquarium World

Mathematics

1 answer:

AlladinOne [14]2 years ago

3 0

Answer:

•the range of packages shipped to rare vinyl is greater than the range of packages shipped to aquarium world.•

•The distribution of the data for packages shipped to both business is approximately symmetrical.•

Step-by-step explanation:

I took the test

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Is it aaaaaaaa or dddddddd

nika2105 [10]

Its none
because a triangle makes up 180 degrees hope this helps and im not a dumby

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3 years ago

Read 2 more answers

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

The sine of angle theta is positive

Remember that

$sin^{2}(\theta)+ cos^{2}(\theta)=1$

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

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2 years ago

a map has a scale of 2in: 6mi. if a two twons on a map is 10 in apart then how far apart are the real twons

kakasveta [241]

Answer:

Step-by-step explanation:

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1 year ago

Paul has 2/3 as many postcards as shawn . the number of postcards shawn has is 3/5 of the number of postcards tim has. if the th

kakasveta [241]

P = ( 2 / 3 )s;
s = ( 3 / 5 )t ;
p + s + t = 280 ;
Then, p =( 2 / 3 )( 3 / 5 )t ;
p = ( 2 / 5 )t;
We solve the equation : ( 2 / 5 )t + ( 3 / 5 )t + t = 280 ;
( 2 / 5 )t + ( 3 / 5 )t + ( 5 / 5 )t = ( 1400 / 5 )t;
10t = 1400 ;
t = 1400 ÷ 10 ;
t = 140 postcards has Tim;
p = ( 2 / 5 ) × 140 = 56 postcards has Paul;
s = ( 3 / 5 ) × 140 = 84 postcards has Shawn ;

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3 years ago

How does the image triangle compare to the pre-image triangle?

BartSMP [9]

A triangle undergoes a sequence of transformations.

First, the triangle is dilated by a scale factor of 1/3 about the origin. Then the triangle is reflected over the x-axis. Finally, the triangle is translated left 3 units and up 2 units.

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1 year ago