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4 years ago

What are the soultions of 3(x-4) (2x-3)= 0 check all

Mathematics

1 answer:

kondaur [170]4 years ago

3 0

Answer:

{3/2, 4}

Step-by-step explanation:

Set x - 4 = 0 and solve for x: x = 4 is a solution.

Seet 2x - 3 = 0 and solve for x: 3/2 is another solution.

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How many solutions do y=3x-4 y=-2x+7

Margaret [11]

Answer:

One.

Step-by-step explanation:

It is a linear equation. This means that x is raised to the first power, which there can only be one solution.

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3 years ago

Read 2 more answers

Tom bought a bicycle on hire purchase from a store. He made a down payment of $250.00 and paid $20.75 every month for a year . H

AlexFokin [52]

Answer:

$499

Step-by-step explanation:

He payed at first a down payment = 250

and 20.75 a month for a year

and there is 12 months in a year which means he payed 20.75 12 times multiply these numbers 12x20.75

=$249

Plus the down payment

=249+250

=<u>$499</u>

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2 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

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

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