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3 years ago

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

2 answers:

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

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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, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

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

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

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 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 z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/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(x + π) = -cos(x) and sin(x + π) = -sin(x),

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

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

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

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