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salantis [7]
4 years ago
8

How do I expand the expression shown?

Mathematics
1 answer:
bezimeni [28]4 years ago
7 0

x² - x + 4x - 4 → D

Expand using the FOIL method

F → multiply First terms → x and x = x²

O → multiply the Outer terms → - 1 and x = - x

I → multiply the Inner terms → 4 and x = 4x

L → multiply the last terms → 4 and - 1 = - 4


(x + 4)(x - 1) = x² - x + 4x - 4


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Which expression is equivalent to 144 Superscript three-halves? 216 1,728 RootIndex 3 StartRoot 12 EndRoot RootIndex 3 StartRoot
Afina-wow [57]

Question:

Which expression is equivalent to 144^(3/2)

Answer:

1728

Step-by-step explanation:

The options are not well presented. However, this is the solution to the question.

Given:

144^(3/2)

Required:

Find Equivalent.

We start my making use of the following law of logarithm.

A^(m/n) = (A^m)^1/n

So,

144^(3/2) = (144³)^½

Another law of indices is that

A^½ = √A

So,

144^(3/2) = (144³)^½ = √(144³)

144³ can be expanded as 144 * 144 * 144.

This gives

144^(3/2) = √(144 * 144 * 144)

The square root can then be splitted to

144^(3/2) = √144 * √144 * √144

144^(3/2) = 12 * 12 * 12

144^(3/2) = 1728.

Hence, the equivalent of 144^(3/2) is 1728

8 0
3 years ago
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15¾÷1¾=?<br><br> O A<br> O B. 9<br> O C. 15¾<br> O D.4¾
Lyrx [107]

Answer:

9

Step-by-step explanation:

first, turn both fractions into improper

turn it into multiplication

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8 0
2 years ago
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All natural numbers are members of which other subsets of real numbers?
vesna_86 [32]
Wholes, integers, rationals, reals
6 0
4 years ago
Nora wants to put new glass in her window
Xelga [282]

Answer:

1632 inches squared

Step-by-step explanation:

Hi!

So the formula for area is length times width.

So you would multiply the length by the width which is 34 times 48 to get the area which is 1632 in squared.

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5 0
4 years ago
Use the Divergence Theorem to evaluate the following integral S F · N dS and find the outward flux of F through the surface of t
Xelga [282]

Answer:

Result;

\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = 32\pi

Step-by-step explanation:

Where:

F(x, y, z) = 2(x·i +y·j +z·k) and

S: z = 0, z = 4 -x² - y²

For the solid region between the paraboloid

z = 4 - x² - y²

div F        

For S: z = 0, z = 4 -x² - y²

We have the equation of a parabola

To verify the result for F(x, y, z) = 2(x·i +y·j +z·k)

We have for the surface S₁ the outward normal is N₁ = -k and the outward normal for surface S₂ is N₂ given by

N_2 = \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{\sqrt{4x^2+4y^2+1} }

Solving we have;

\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = \int\limits\int\limits_{S1} { \textbf{F}} \, \cdot \textbf{N}_1 d {S} + \int\limits\int\limits_{S2} { \textbf{F}} \, \cdot \textbf{N}_2 d {S}

Plugging the values for N₁ and N₂, we have

= \int\limits\int\limits_{S1} { \textbf{F}} \, \cdot \textbf{(-k)}d {S} + \int\limits\int\limits_{S2} { \textbf{F}} \, \cdot  \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{\sqrt{4x^2+4y^2+1} } d {S}

Where:

F(x, y, z) = 2(xi +yj +zk) we have

= -\int\limits\int\limits_{S1} 2z \ dA + \int\limits\int\limits_{S2} 4x^2+4y^2+2z \ dA

= -\int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 2z \ dA + \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2+2z \ dA

= \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2 \ dxdy

= \int\limits^2_{-2} \frac{(16y^2 +32)\sqrt{-(y^2-4)} }{3} dy

= 32π.

6 0
4 years ago
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