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3 years ago

What is the solution of |2x| = –4?

Mathematics

1 answer:

DENIUS [597]3 years ago

7 0

-2x=-4 or x=2
2x =-4 or x=-2

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Find the midpoint of the line segment shown below.

balu736 [363]

Answer:

x²-6x+y²-4y=-4

Step-by-step explanation:

We know the circle equation stands like this:

(x-a)² + (y-b)² = R²

and (a,b) is center

and R stands for Radius

so in this case:

center = (3,2)

R = 3

(x-3)² + (y-2)² = 9

x²-6x+9 + y²-4y+4=9

x²-6x+y²-4y=-4

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2 years ago

Find the value of x. B. X+2 3 E Х 2 С А x = [?]

Alik [6]

Answer:

The simplest form of the function is f(x) = x2. The graph is a parabola often called the basic parabola. Notice that the graph is symmetrical about the y- axis. The y- axis is called the axis of symmetry of this function

8 0

2 years ago

Which statement is NOT true about this figure?

Lunna [17]

Answer:

od I'd the correct answer

8 0

2 years ago

A candle is in the shape of a square pyramid. Its height is 5 cm and its slant height is 8 cm

Akimi4 [234]

Using the Pythagoras theorem

8^2 = 5^2 + x^2 where x = 1/2 length of the base

x^2 = 64-25 = 39

x = = sqrt39 = 6.24

so base length = 2*6.24 = 12.5 cm to nearest tenth

3 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