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2 years ago

What is the distance from the origin to point A graphed on the complex plane below?

Mathematics

1 answer:

Kobotan [32]2 years ago

5 0

Answer:

√13

Step-by-step explanation:

(-3, -2) (0, 0)

distance √(9 + 4) = √13

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Tom swims a 1/2 kilometers every 1/4 hour. How far will he swim in one hour.

fgiga [73]

Answer:

2 kilometers

Step-by-step explanation:

every 1 kilometer is a 1/2 hour. double that and you get 2

8 0

2 years ago

Solve the given proportion for x. Answer to two decimal places. x/5=16/21

Angelina_Jolie [31]

$\\ \sf\longmapsto \dfrac{x}{5}=\dfrac{16}{21}$

$\\ \sf\longmapsto 21x=16(5)$

$\\ \sf\longmapsto 21x=80$

$\\ \sf\longmapsto x=\dfrac{80}{21}$

$\\ \sf\longmapsto x=3.6$

3 0

2 years ago

Read 2 more answers

Please help me out with this one

Vlad [161]

Might be late but the first one is 125 and the seced one is less than :)

4 0

2 years ago

13 x-2(8+5 x)=12-11 x

77julia77 [94]

Expand

13x - 16 - 10x = 12 - 11x

Simplify 13x - 16 - 10x to 3x - 16

3x - 16 = 12 - 11x

Add 16 to both sides

3x = 12 - 11x + 16

Simplify 12 - 11x + 16 to -11x + 28

3x = -11x + 28

Add 11x to both sides

3x + 11x = 28

Simplify 3x + 11x to 14x

14x = 28

Divide both sides by 14

x = 28/14

Simplify 28/14 to 2

x = 2

4 0

2 years ago

What's the volume? I honestly just need the formula.

Arturiano [62]

Answer:

$v=364.5\ m^3$

Step-by-step explanation:

Volume Of A Regular Solid

When a solid has a constant cross-section, the volume can be found by multiplying the area of the base by the height. The area of a trapezium is

$\displaystyle A_t=\frac{b_1+b_2}{2}h$

where $b_1$ and $b_2$ are the lengths of the parallel sides and h the distance between them.

The figure shows a solid with a trapezoid as the constant cross-section and a height x. The volume of the solid is

$\displaystyle v=A_t\ x$

$\displaystyle v=\frac{b_1+b_2}{2}\ h\ x$

The image doesn't explicitly say if the length of 4.5 is the height of the trapezium or the length of that side. We'll assume the first, so our data is:

$\displaystyle b_1=7m,\ b_2=11m,\ h=4.5m,\ x=9m$

We now compute the volume

$\displaystyle v=\frac{7+11}{2}.(4.5)(9)=364.5$

$\boxed{\displaystyle v=364.5\ m^3}$

8 0

3 years ago