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allsm [11]
3 years ago
11

P•X=-10 I already know the answer but what why is P -10?

Mathematics
2 answers:
Yuri [45]3 years ago
7 0
P-x= -10
Now to isolate x divide both sides by p
-x = -10/p
But x can not be a negative do distribute a negative 1 to both sides and your answer is :
x = 10/p
strojnjashka [21]3 years ago
3 0
P•x = - 10
so in conclusion
x=10/p
You might be interested in
Which is bigger square root of 1001 - square root of 1000 or 1/10?
IgorC [24]
Square root of 1001=10.0033
square root of 1000=10
so 10.0033-10=0.0033
1/10=0.1
so, 1/10 is bigger than square root of 1001- square root of 1000
8 0
3 years ago
If the volume of a sphere inscribed in a cubic box is 36pi, find the surface area of the box
Kruka [31]

Answer:

Surface area of the cube = 216 square units

Step-by-step explanation:

Let the length of a side of a cube = x unit

Diameter of the sphere inscribed in this cube = length of a side of the cube

Volume of the sphere = \frac{4}{3}\pi r^{3}

Where r = radius of the sphere = \frac{x}{2} units

36π = \frac{4}{3}\pi (\frac{x}{2})^{3}

36π = \frac{4x^{3}\pi }{24}

36×24 = 4x³

x = \sqrt[3]{\frac{36\times 24}{4} }

x = 6 units

Length of a side of the cube = 6 units

Surface area of the cube = 6×(Side)²

                                          = 6×(6)²

                                          = 216

                                          = 216 square units    

3 0
3 years ago
5. Consider the following equations.
WITCHER [35]

Answer:

Step-by-step explanation:

In y = mx + b form, the slope can be found in the m position and the y intercept can be found in the b position.

4x - 2y = -6

-2y = -4x- 6

y = (-4/-2)x -(6/-2)

y = 2x + 3....slope here is 2, and y int is 3

3x + y = 3

y = -3x + 3.....slope here is -3, and y int is 3

B. slopes are different, y intercepts are the same

3 0
3 years ago
Select the correct answer.
Digiron [165]
C because if you plug one into the other the answer comes out as x and that is how you know it is an inverse
7 0
2 years ago
Let z=3+i, <br>then find<br> a. Z²<br>b. |Z| <br>c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq
zysi [14]

Given <em>z</em> = 3 + <em>i</em>, right away we can find

(a) square

<em>z</em> ² = (3 + <em>i </em>)² = 3² + 6<em>i</em> + <em>i</em> ² = 9 + 6<em>i</em> - 1 = 8 + 6<em>i</em>

(b) modulus

|<em>z</em>| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(<em>z</em>) = arctan(1/3)

Then

<em>z</em> = |<em>z</em>| exp(<em>i</em> arg(<em>z</em>))

<em>z</em> = √10 exp(<em>i</em> arctan(1/3))

or

<em>z</em> = √10 (cos(arctan(1/3)) + <em>i</em> sin(arctan(1/3))

(c) square root

Any complex number has 2 square roots. Using the polar form from part (d), we have

√<em>z</em> = √(√10) exp(<em>i</em> arctan(1/3) / 2)

and

√<em>z</em> = √(√10) exp(<em>i</em> (arctan(1/3) + 2<em>π</em>) / 2)

Then in standard rectangular form, we have

\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)

and

\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)

We can simplify this further. We know that <em>z</em> lies in the first quadrant, so

0 < arg(<em>z</em>) = arctan(1/3) < <em>π</em>/2

which means

0 < 1/2 arctan(1/3) < <em>π</em>/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}

\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}

and since cos(<em>x</em> + <em>π</em>) = -cos(<em>x</em>) and sin(<em>x</em> + <em>π</em>) = -sin(<em>x</em>),

\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}

\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}

Now, arctan(1/3) is an angle <em>y</em> such that tan(<em>y</em>) = 1/3. In a right triangle satisfying this relation, we would see that cos(<em>y</em>) = 3/√10 and sin(<em>y</em>) = 1/√10. Then

\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}

\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}

\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}

\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}

So the two square roots of <em>z</em> are

\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}

and

\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}

3 0
3 years ago
Read 2 more answers
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