All categories

1 year ago

Solve u^2 = -4, where u is a real number. Simplify your answer as much as possible.

Mathematics

1 answer:

alexdok [17]1 year ago

8 0

Answer:

2i and -2i

Step-by-step explanation:

I'll change u to x to be easier to understand.

x^2 = -4

sqrt(x^2) = sqrt(-4)

sqrt(-4) = 2i, or -2i

You might be interested in

Need help for question 8 and 9. I need the equation that’s passes through the points!

dsp73

Answer:

Step-by-step explanation:

Question 8

Let the equation of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by,

y = mx + b

Here, m = slope and y-intercept = b

Since, m = $\frac{y_2-y_1}{x_2-x_1}$

For two points (1, 4) and (5, 8),

m = $\frac{8-4}{5-1}$

m = 1

Equation will be,

y = (1)x + b

y = x + b

Since, this line passes through (1, 4),

4 = 1 + b

b = 3

Therefore, equation of the line will be,

y = x + 3

Question 9

Let the equation is y = mx + b

m = $\frac{y_2-y_1}{x_2-x_1}$

For two points (2, 10) and (6, 4)

m = $\frac{10-4}{2-6}$

m = $-\frac{3}{2}$

By substituting the value of 'm' in the equation,

y = $-\frac{3}{2}x+b$

Since, this line passes through (2, 10)

10 = $-\frac{3}{2}(2)+b$

b = 10 + 2

b = 12

Therefore, equation of the line will be,

$y=-\frac{3}{2}x+12$

4 0

3 years ago

N is a whole number such that 6< 3n < 15 List all the possible values of n.

Sav [38]

Answer:

3, 4

Step-by-step explanation:

6 < 3n < 15

Divide the three "sides" by 3.

2 < n < 5

n must be between 2 and 5,a nd it must be a whole number.

n can be 3 or 4

7 0

2 years ago

Read 2 more answers

Look at the following numbers: -5, -6, 0, 6

gladu [14]

C is the answer l.

6 + (-6) = 0

4 0

2 years ago

Read 2 more answers

Which is bigger -1 1/2 or -1/2

bixtya [17]

Answer:

-1/2

Step-by-step explanation:

just trust me, idk how to explain it

5 0

3 years ago

Find all the values of x in the set of complex numbers that satisfy the following equation:

Fynjy0 [20]

Compute all the component integrals first:

$I_1=\displaystyle\int_0^{\pi/4}2\sec x\,\mathrm dx=2\ln(\sqrt2+1)$
$I_2=\displaystyle\int_0^2\ln x\,\mathrm dx=2(\ln2-1)$
$I_3=\displaystyle\lim_{a\to\infty}\int_{-a}^a\frac{\mathrm dx}{x^2+1}=\pi$

Now,

$\sqrt2\approx1.4\implies \sqrt2+1\approx2.4$
$\implies \left\lceil I_1\right\rceil=2$

$e$
$\implies\left\lfloor I_2\right\rfloor=-1$

$\pi\approx3.14\implies\left\lceil I_3\right\rceil=4$

So the given equation reduces to

$\displaystyle\sum_{k=-1}^2\frac{\mathrm d}{\mathrm dx}x^{k+2}=1-4!$
$\dfrac{\mathrm dx}{\mathrm dx}+\dfrac{\mathrm dx^2}{\mathrm dx}+\dfrac{\mathrm dx^3}{\mathrm dx}+\dfrac{\mathrm dx^4}{\mathrm dx}=-23$
$4x^3+3x^2+2x+24=0$

a fairly standard cubic. Incidentally, when $x=-2$ , the LHS reduces to 0, so $x+2$ is a factor of the cubic. You can find the remaining two solutions easily with the quadratic formula.

5 0

2 years ago