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jasenka [17]
3 years ago
14

What's the difference between 6z and z6

Mathematics
1 answer:
Airida [17]3 years ago
4 0
Simple..

there is a major difference between 6z and z^{6}

6z means: 6 times z

and

z^{6} means: z to the power of 6(z*z*z*z*z*z)

An example...

make z=2...plug n chug..

6z----> 6(2)=12
and
z^{6}--->2^{6}=64

As you can see..there is a major difference.

Thus, your answer.
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Show your solution follow the steps need an answer, please need an answer, please.​
STatiana [176]

Answer:

Step-by-step explanation:

27 ) 2x² - x - 1 = 0

2x² - 2x +  x  - 1 = 0

2x ( x - 1) + (x - 1) = 0

(x - 1)(2x + 1) = 0

x - 1 = 0  ; 2x + 1 = 0

x = 1   ;     2x = -1

                  x = -1/2

x = 1 ; -1/2

Option b

28) Area of a rectangle = 24 cm²

length * width = 24

(3x +2 )(2x -1) = 24

3x(2x -1) + 2(2x - 1) = 24

3x *2x - 3x *1 + 2*2x - 2*1 = 24

6x² -3x + 4x - 2= 24

6x² + x - 2 -24 = 0

6x² + x - 26 = 0

6x² -12x + 13x - 26 = 0

6x(x - 2) + 13(x - 2) = 0

(x -2)(6x +13) = 0

x = 2      {Ignore 6x + 13 as it gives negative value}

length = 3x + 2 = 3*2 + 2 = 6 + 2 = 8 cm

Width  = 2x - 1 = 2*2 - 1 = 4 - 1 = 3 cm

29) Area of square = 900 cm²

Side² = 900

(5x)² = 900

25x² = 900

x² = 900/25

x² = 36

x = √36 = √6*6

x = 6 cm

30) base = b cm

height = b + 2

Area of triangle =  24 cm²

\dfrac{1}{2}base*height = 24\\\\\dfrac{1}{2}*b*(b+2) = 24\\

b(b + 2) = 24*2

b² + 2b = 48

b² + 2b - 48 = 0

b²  - 6b + 8b  - 48 = 0

b(b - 6) + 8(b - 6) = 0

(b - 6) (b + 8) = 0

b - 6 = 0    {Ignore b +8 = 0 as it gives negative value}

b = 6 cm

height = 6+ 2 = 8 cm

5 0
3 years ago
I need help finding the values of the last boxes shown in the image.
Bingel [31]

The volume of the region R bounded by the x-axis is: \mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

<h3>What is the volume of the solid revolution on the X-axis?</h3>

The volume of a solid is the degree of space occupied by a solid object. If the axis of revolution is the planar region's border and the cross-sections are parallel to the line of revolution, we may use the polar coordinate approach to calculate the volume of the solid.

In the graph, the given straight line passes through two points (0,0) and (2,8).

Therefore, the equation of the straight line becomes:

\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}

where:

  • (x₁, y₁) and (x₂, y₂) are two points on the straight line

Thus, from the graph let assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8), we have:

\mathbf{y-0 = \dfrac{8-0}{2-0}(x-0)}

y = 4x

Now, our region bounded by the three lines are:

  • y = 0
  • x = 2
  • y = 4x

Similarly, the change in polar coordinates is:

  • x = rcosθ,
  • y = rsinθ

where;

  • x² + y² = r²  and dA = rdrdθ

Now

  • rsinθ = 0   i.e.  r = 0 or θ = 0
  • rcosθ = 2 i.e.   r  = 2/cosθ
  • rsinθ = 4(rcosθ)  ⇒ tan θ = 4;  θ = tan⁻¹ (4)

  • ⇒ r = 0   to   r = 2/cosθ
  •    θ = 0  to    θ = tan⁻¹ (4)

Then:

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

Learn more about the determining the volume of solids bounded by region R here:

brainly.com/question/14393123

#SPJ1

5 0
2 years ago
How much would 180 pens cost if bought in boxes of 12, each box costing $5.80?
harina [27]
87 DOLLARS IS THE ANSWER

7 0
4 years ago
Read 2 more answers
Question
Gnom [1K]

Answer:

ask to your teachers and you will be finish the work

3 0
2 years ago
Use spherical coordinates. evaluate (9 − x2 − y2) dv, where h is the solid hemisphere x2 + y2 + z2 ≤ 4, z ≥ 0.
avanturin [10]
In spherical coordinates, we have

\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}

which gives volume element

\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta

and so the triple integral is given by

\displaystyle\iiint_H(9-x^2-y^2)\,\mathrm dV
=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2}\int_{\rho=0}^{\rho=2}(9-\rho^2\sin^2\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta
=\displaystyle2\pi\int_{\varphi=0}^{\varphi=\pi/2}\int_{\rho=0}^{\rho=2}(9\rho^2\sin\varphi-\rho^4\sin^3\varphi)\,\mathrm d\rho\,\mathrm d\varphi
=\dfrac{592\pi}{15}
4 0
4 years ago
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