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4 years ago

Given 2x² = -8, x is

Mathematics

2 answers:

DanielleElmas [232]4 years ago

8 0

Answer:

2x²=-8

x²= 8/2

x²= 4

x= √4

x= 2

So, the value of x is 2

nlexa [21]4 years ago

4 0

Answer:

its 2i or -2i

Step-by-step explanation:

just did it and also the next one is -4 and 10

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Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

PLEASE HELP!!!!!!!!! ASAP!!!!! I REWARD BRAINLIEST!!!! HAVE CLASS IN 30 MINUTES!1111

KiRa [710]

Answer:

$y = \dfrac{1}{3}x + 1$

Step-by-step explanation:

The given line is in slope-intercept form,

y = mx + b,

where m is the slope.

The slope of the given line is 1/3, so m = 1/3.

Parallel lines have equal slopes, so the slope of the parallel line is also 1/3.

y = 1/3 x + b

Now we can find the equation of the parallel line through point (6, 3) by using the given point's coordinates for x and y and solving for b.

3 = (1/3)(6) + b

3 = 2 + b

b = 1

Equation: $y = \dfrac{1}{3}x + 1$

8 0

3 years ago

Read 2 more answers

Can any negative integer be greater than a positive integer explain

Keith_Richards [23]

No, it is impossible. Intuitively, a negative number sits at the left of 0 on the number line, and a positive number sits at the right of 0 on the number line. And a number x is greater than another number y if x sits at the right of y on the number line. So, every positive number is greater than any negative number.

Also, by definition, a positive number is greater than 0, and a negative number is smaller than zero. So, if x is positive and y is negative, you have

$y < 0 < x$