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3 years ago

Find the sum of (3 – 4i) and (6 + 7i)

Mathematics

1 answer:

Ivanshal [37]3 years ago

7 0

Answer:-(3-4i) and -(6+7i)

Step-by-step explanation: To find the complex conjugate, negate the term with i.

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15. Find the force on an electron crossing a uniform magnetic field of intensity 0.5 T WILII

EleoNora [17]

Answer:

F = 8.48 x $10^{-18}$ N

Step-by-step explanation:

The force on an electron in a magnetic field can be determined by;

F = qvB Sin θ

where: e is the electron, v is its velocity, B is the measure of the magnetic field, and θ is the angle of its path.

Thus,

q = -1.6 x $10^{-19}$ C

v = 106 m/s

B = 0.5 T

θ = $90^{o}$

So that,

F = 1.6 x $10^{-19}$ x 106 x 0.5

= 8.48 x $10^{-18}$ N

The force on the electron crossing the uniform magnetic field is 8.48 x $10^{-18}$ N.

8 0

3 years ago

You cut 63 meters of rope into 9 equal pieces. How long is each equal piece of rope?

nirvana33 [79]

Each piece of rope will be 7 meters

5 0

3 years ago

Read 2 more answers

Hitunglah nilai x ( jika ada ) yang memenuhi persamaan nilai mutlak berikut . Jika tidak ada nilai x yang memenuhi , berikan ala

Julli [10]

(a). The solutions are 0 and ⁸/₃.

(b). The solutions are 1 and ¹³/₃.

(c). The equation has no solution.

(d). The only solution is ²¹/₂₀.

(e). The equation has no solution.

<h3>Further explanation</h3>

These are the problems with the absolute value of a function.

For all real numbers x,

$\boxed{ \ |f(x)|=\left \{ {{f(x), for \ f(x) \geq 0} \atop {-f(x), for \ f(x) < 0}} \right. \ }$

Problem (a)

|4 – 3x| = |-4|

|4 – 3x| = 4

Case 1

$\boxed{ \ 4 - 3x \geq 0 \ } \rightarrow \boxed{ \ 4\geq 3x \ } \rightarrow \boxed{ \ x\leq \frac{4}{3} \ }$

For 4 – 3x = 4

Subtract both sides by four.

-3x = 0

Divide both sides by -3.

x = 0

Since $\boxed{ \ 0\leq \frac{4}{3} \ }$ , x = 0 is a solution.

Case 2

$\boxed{ \ 4 - 3x < 0 \ } \rightarrow \boxed{ \ 4 < 3x \ } \rightarrow \boxed{ \ x > \frac{4}{3} \ }$

For -(4 – 3x) = 4

-4 + 3x = 4

Add both sides by four.

3x = 8

Divide both sides by three.

$x = \frac{8}{3}$

Since $\boxed{ \ \frac{8}{3} > \frac{4}{3} \ }$ , $\boxed{ \ x = \frac{8}{3} \ }$ is a solution.

Hence, the solutions are $\boxed{ \ 0 \ and \ \frac{8}{3} \ }$

————————

Problem (b)

2|3x - 8| = 10

Divide both sides by two.

|3x - 8| = 5

Case 1

$\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }$

For 3x - 8 = 5

Add both sides by eight.

3x = 13

Divide both sides by three.

$x = \frac{13}{3}$

Since $\boxed{ \ \frac{13}{3} \geq \frac{4}{3} \ }$ , $\boxed{ \ x = \frac{13}{3} \ }$ is a solution.

Case 2

$\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }$

For -(3x – 8) = 5

-3x + 8 = 5

Subtract both sides by eight.

-3x = -3

Divide both sides by -3.

x = 1

Since $\boxed{ \ 1 < \frac{8}{3} \ }$ , $\boxed{ \ x = 1 \ }$ is a solution.

Hence, the solutions are $\boxed{ \ 1 \ and \ \frac{13}{3} \ }$

————————

Problem (c)

2x + |3x - 8| = -4

Subtracting both sides by 2x.

|3x - 8| = -2x – 4

Case 1

$\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }$

For 3x – 8 = -2x – 4

3x + 2x = 8 – 4

5x = 4

$x = \frac{4}{5}$

Since $\boxed{ \ \frac{4}{5} \ngeq \frac{8}{3} \ }$ , $\boxed{ \ x = \frac{4}{5} \ }$ is not a solution.

Case 2

$\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }$

For -(3x - 8) = -2x – 4

-3x + 8 = -2x – 4

2x – 3x = -8 – 4

-x = -12

x = 12

Since $\boxed{ \ 12 \nless \frac{8}{3} \ }$ , $\boxed{ \ x = 12 \ }$ is not a solution.

Hence, the equation has no solution.

————————

Problem (d)

5|2x - 3| = 2|3 - 5x|

Let’s take the square of both sides. Then,

[5(2x - 3)]² = [2(3 - 5x)]²

(10x – 15)² = (6 – 10x)²

(10x - 15)² - (6 - 10x)² = 0

According to this formula $\boxed{ \ a^2 - b^2 = (a + b)(a - b) \ }$

$[(10x - 15) + (6 - 10x)][(10x - 15) - (6 - 10x)]] = 0$

(-9)(20x - 21) = 0

Dividing both sides by -9.

20x - 21 = 0

20x = 21

$x = \frac{21}{20}$

The only solution is $\boxed{ \ \frac{21}{20} \ }$

————————

Problem (e)

2x + |8 - 3x| = |x - 4|

We need to separate into four cases since we don’t know whether 8 – 3x and x – 4 are positive or negative. We cannot square both sides because there is a function of 2x.

Case 1

8 – 3x is positive (or 8 - 3x > 0)
x – 4 is positive (or x - 4 > 0)

2x + 8 – 3x = x – 4

8 – x = x – 4

-2x = -12

x = 6

Substitute x = 6 into 8 – 3x ⇒ 8 – 3(6) < 0, it doesn’t work, even though when we substitute x = 6 into x - 4 it does work.

Case 2

8 – 3x is positive (or 8 - 3x > 0)
x – 4 is negative (or x - 4 < 0)

2x + 8 – 3x = -(x – 4)

8 – x = -x + 4

x – x = = 4 - 8

It cannot be determined.

Case 3

8 – 3x is negative (or 8 - 3x < 0)
x – 4 is positive. (or x - 4 > 0)

2x + (-(8 – 3x)) = x – 4

2x – 8 + 3x = x - 4

5x – x = 8 – 4

4x = 4

x = 1

Substitute x = 1 into 8 - 3x, $\boxed{ \ 8 - 3(1) \nless 0 \ }$ , it doesn’t work. Likewise, when we substitute x = 1 into x – 4, $\boxed{ \ 1 - 4 \not> 0 \ }$

Case 4

8 – 3x is negative (or 8 - 3x < 0)
x – 4 is negative (or x - 4 < 0)

2x + (-(8 – 3x)) = -(x – 4)

2x – 8 + 3x = -x + 4

5x + x = 8 – 4

6x = 4

$\boxed{ \ x=\frac{4}{6} \rightarrow x = \frac{2}{3} \ }$

Substitute $x = \frac{2}{3} \ into \ 8-3x$ , $\boxed{ \ 8 - 3 \bigg(\frac{2}{3}\bigg) \not< 0 \ }$ , it doesn’t work. Even though when we substitute $x = \frac{2}{3} \ into \ x-4$ , $\boxed{ \ \bigg(\frac{2}{3}\bigg) - 4 < 0 \ }$ it does work.

Hence, the equation has no solution.

<h3>Learn more</h3>

The inverse of a function brainly.com/question/3225044
The piecewise-defined functions brainly.com/question/9590016
The composite function brainly.com/question/1691598

Keywords: hitunglah nilai x, the equation, absolute value of the function, has no solution, case, the only solution

5 0

3 years ago

Read 2 more answers

Given g(x) √x-4 and h(x) = 2x - 8

RoseWind [281]

Answer:

Step-by-step explanation:

Given : g(x) = and h(x)=2x-8.

g*h(x) =(2x-8).

We have square root(x-6) in composite function f*h(x).

So, we need to find the domain, we need to check for that values of x's, square root(x-4\6) would be defined.

Square roots are undefined for negative values.

Therefore, we can setup an inequality for it's domain x-6≥0.

Adding 4 on both sides, we get

x-6+6≥0+6.

x≥6.

8 0

3 years ago

Read 2 more answers

Matlab the equation of a circle with its center at x=3 and y=2 is given by , where r is the radius of the circle. first derive t

Blizzard [7]

The equation of a circle with center at (h,k) and radius r is
$(x-h)^2+(y-k)^2=r^2$
we are given
center is at (3,2)
$(x-3)^2+(y-2)^2=r^2$
solving for y
$(y-2)^2=-1(x-3)^2+r^2$
$y-2=\sqrt{-1(x-3)^2+r^2}$
$y=2+\sqrt{-1(x-3)^2+r^2}$
take the derivitive to find the slope at any point
$\frac{dy}{dx}=((-1)(x-3)^2+r^2))(-x+3)$

we can use point slope form
for apoint (h,k) and slope m, the equation is
y-k=m(x-h)
not sure if you want in terms of what
I'll just say for a point (h,k)
so we get
$y-k=((-1)(x-3)^2+r^2))(-x+3)(x-h)$
where $k=2+\sqrt{-1(h-3)^2+r^2}$
that's the equation of the tangent line at (h,k) for arbitrary values of r

do the plots and other stuff yourself

7 0

3 years ago