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3 years ago

Calculate the distance between (-2 - 5i) and (1 + 3i)

Mathematics

1 answer:

Andru [333]3 years ago

4 0

Answer:

d = sqrt(73)

d is approximately 8.544003745

Step-by-step explanation:

To find the distance we use the formula

d = sqrt ( (x2 -x1)^2 + (y2-y1)^2)

where x are the real coordinates and y is the imaginary coordinates

d = sqrt ( (1--2)^2 + (3--5)^2)

d = sqrt ( (1+2)^2 + (3+5)^2)

d = sqrt ( (3)^2 + (8)^2)

d = sqrt( 9+64)

d = sqrt(73)

d is approximately 8.544003745

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What properties are used in 2(3x+5)=22+5x-3x

finlep [7]

2(3x+5)=22+5x-3x
using associative property we shall open the parenthesis
6x+10=22+5x-3x
6x+10=22+2x
using cummutative property and additive identity property we shall put like terms together
6x-2x=22-10
4x =12
x=12/4
x=3

7 0

4 years ago

Given that a trapezoid has the following dimensions, find the area. Area = ________ units^2 (only put the number)

patriot [66]

Trapezoid area = ((sum of the bases) ÷ 2) • height

Trapezoid area = ((5 + 3) / 2) * 5

Trapezoid area = ((8) / 2) *5

Trapezoid area = 20

Source:
http://www.1728.org/quadtrap.htm

5 0

4 years ago

Help me please algebra 2 PLEASSEEEEE!!!

liq [111]

Answer:

29.74 degrees.

Step-by-step explanation:

Tan(theta) = opposite / adjcent.

opposite = 8

adjacent = 14

Tan(theta) = 8/14

Tan(theta) = 0.5714

theta = tan^-1(0.5714)

theta = 29.74 degrees

5 0

4 years ago

Name an equivalent ratio for 1/2 with a denominator of 20

soldi70 [24.7K]

1/2=x/20
times both sides by 20
20/2=x
10=x
answer is 10/20

5 0

3 years ago

Read 2 more answers

Determine the equation of the line that passes through the points of intersection of the graphs of the quadratic functions f(x)

timofeeve [1]

Answer: $x-2y-2=0$

Step-by-step explanation:

Given :

$f(x) = x^2 - 4 \\ g(x) = - 3x^2 + 2x + 8$

Point of intersection :

$f(x)=g(x)\\x^2-4=-3x^2+2x+8\\4x^2-2x-12=0\\2x^2-x-6=0\\2x^2-4x+3x-6=0\\2x(x-2)+3(x-2)=0\\(x-2)(2x+3)=0\\x=2\,,\,\frac{-3}{2}$

$x=2\,;f(2)=2^2-4=0\\x=\frac{-3}{2}\,; f\left ( \frac{3}{2} \right )=\left ( \frac{3}{2} \right )^2-4=\frac{-7}{4}=-1.75$

So, we have points $\left ( 2,0 \right )\,,\,\left ( -1.5,-1.75\ \right )$

Equation of line passing through two points $\left ( x_1,y_1 \right )\,,\,\left ( x_2,y_2 \right )$ is given by $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left ( x-x_1 \right )$

Let $\left ( x_1,y_1 \right )=\left ( 2,0 \right )\,,\,\left ( x_2,y_2 \right )=\left ( -1.5,-1.75\ \right )$

So, equation is as follows :

$y-0=\frac{-1.75-0}{-1.5-2}\left ( x-2 \right )\\y=\frac{-1.75}{-3.5}\left ( x-2 \right )\\y=\frac{1}{2}(x-2)\\2y=x-2\\x-2y-2=0$

4 0

4 years ago