All categories

3 years ago

What is the square root of 75 using imaginary numbers

Mathematics

1 answer:

Solnce55 [7]3 years ago

5 0

Answer:

8.660254038

Step-by-step explanation:

or 5✓3

hope this helps

You might be interested in

nicole used 24 pieces of fruit .if 1/6 of them were peaches how many peaches in all did janeen and nicole use to make their frui

Rzqust [24]

They used 4 peaches because if 1/6 of the 24 fruits were peaches then there were 4 peaches.

8 0

4 years ago

Read 2 more answers

1) Para que valores reais de x o ponto P (3x -6, 2x +4) pertence ao 4º quadrante?

skelet666 [1.2K]

Answer:

There are no real values of x for point P to belong to the 4 quadrant

Step-by-step explanation:

<u><em>The question in English is</em></u>

To what real values of x does the point P (3x -6, 2x +4) belong to the 4th quadrant?

we know that

A point in the fourth Quadrant has the x-coordinate positive and the y-coordinate negative

we have the point

P (3x-6, 2x+4)

$3x-6 > 0$ ----> inequality A ( x-coordinate must be positive)

$2x+4 < 0$ ---> inequality B ( y-coordinate must be negative)

Solve Inequality A

$3x-6 > 0$

$3x > 6$

$x > 2$ -----> (2,∞)

Solve Inequality B

$2x+4 < 0$

$2x < -4$

$x < -2$ ----> (-∞,-2)

The solution of the system is

(-∞,-2) ∩ (2,∞)

therefore

The system has no solution

There are no real values of x for point P to belong to the 4 quadrant

3 0

4 years ago

What is the area of this trapezoid?

ZanzabumX [31]

area of trapezoid

A=(a+b/2)h

the answer is c 20mm²

6 0

2 years ago

Can anyone do this quick please

evablogger [386]

-3 because -8 plus -2 equal 10 and 6 plus - 1 equal 7 they need -3 to get to 10

3 0

3 years ago

Read 2 more answers

How do you factor 2x^3+5y^3

Pavel [41]

All you do is...
$\mathrm{Apply\:sum\:of\:cubes\:rule:\:}x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)$

$2x^3+5y^3=\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(\left(\sqrt[3]{2}\right)^2x^2-\sqrt[3]{2}\sqrt[3]{5}xy+\left(\sqrt[3]{5}\right)^2y^2\right)$
$\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(\left(\sqrt[3]{2}\right)^2x^2-\sqrt[3]{2}\sqrt[3]{5}xy+\left(\sqrt[3]{5}\right)^2y^2\right) \ \textgreater \ Refine$

$\left(\sqrt[3]{2}x+\sqrt[3]{5}y\right)\left(2^{\frac{2}{3}}x^2-\sqrt[3]{10}xy+5^{\frac{2}{3}}y^2\right)$

Hope this helps!

4 0

3 years ago