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3 years ago

2m/(2m 3)-(2m)/(2m-3)=1how many extraneous solutions does the equation have?

1 answer:

4 0

$\frac{2m}{2m+3} - \frac{2m}{2m-3} =1 \\ \frac{2m(2m-3)-2m(2m+3)}{4m^2-9} =1 \\ \frac{4m^2-6m-4m^2-6m}{4m^2-9} =1 \\ \frac{-12m}{4m^2-9} =1 \\ -12m=4m^2-9 \\ 4m^2+12m-9=0 \\ m=0.6213 \ or \ m=-3.6213$
For m = 0.6213, $\frac{2(0.6213)}{2(0.6213)+3} - \frac{2(0.6213)}{2(0.6213)-3} \\ \frac{1.2426}{4.2426} - \frac{1.2426}{-1.7574} =0.2929+0.7071=1$
For m = -3.6213, $\frac{2(-3.6213)}{2(-3.6213)+3} - \frac{2(-3.6213)}{2(-3.6213)-3} \\ \frac{-7.2426}{-4.2426} - \frac{-7.2426}{-10.2426} =1.7071-0.7071=1$
Therefore, the equation has no extraneous solution.

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Please help soon!

Jlenok [28]

Answer:

a) Translate the graph of f(x) down 3 units.

b) (8,-1) is a point on the graph of g(x)

Step-by-step explanation:

The transformations that subtract a constant number of units from the functional expression f(x) are transformations that lower the graph of the function in those many units,

Therefore they correspond to a translation of the original graph down the number of units involved.

In this case, the number of units involved is "3" (due to the "-3" added to the expression for f(x). So he correct answer for the first part is: Translate the graph of f(x) down 3 units.

For the second part, one has to try each of the coordinate pairs given in the new function g(x) to see which one results in a true statement:

1) Testing (-8,-1) by checking if replacing x with the value "-8" renders "-1" for the y-value: $g(x)=\sqrt[3]{x} -3\\g(-8)=\sqrt[3]{-8} -3\\g(-8)=-2-3\\g(-8)=-5$

so this is NOT a point on the graph of g(x).

2) Testing (-1,-2) by checking if replacing x with the value "-1" renders "-2" for the y-value: $g(x)=\sqrt[3]{x} -3\\g(-1)=\sqrt[3]{-1} -3\\g(-1)=-1-3\\g(-1)=-4$

so this is NOT a point on the graph of g(x).

3) Testing (2,-1) by checking if replacing x with the value "2" renders "-1" for the y-value: $g(x)=\sqrt[3]{x} -3\\g(2)=\sqrt[3]{2} -3\\$

the cubic root of 2 is not a rational number, because 2 is not a perfect cube, so the expression cannot be reduced, so this is NOT a point on the graph of g(x).

4) Testing (8,-1) by checking if replacing x with the value "8" renders "-1" for the y-value: $g(x)=\sqrt[3]{x} -3\\g(8)=\sqrt[3]{8} -3\\g(8)=2-3\\g(8)=-1$

Therefore this pair (8,-1) IS a point on the graph of g(x).

5 0

3 years ago

Read 2 more answers

A jumbo chocolate bar with a rectangular shape measures 11 centimeters in length, 8 centimeters in width, and 3 centimeters in

jolli1 [7]

11ce^3 nt^2 imrs, 8ce^3 nt^2 imrs, 3ce^3 nt^2 imrs, 0.2, 3ce^3 nt^2 imr, ce^3 nt^2 imrs

3 0

3 years ago

Please help its urgent please

elena-s [515]

Answer:

3200000 I think

Step-by-step explanation:

I just multiplied them together.

4 0

3 years ago

HURRY!! PLEASE!!

Marina86 [1]

Answer:

c. y=56(2)x-1 sorry if its wrong

4 0

4 years ago

I need help now plz

Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

The picture is a bit blurry, but from what i can see, i'd say C (whichever one has -4 x 3 x 1 under the square root.)

So for this, you need to know the quadratic formula, which is:

$\frac{-b ± \sqrt{b^{2} -4ac} }{2a}$

*ignore the weird capital A, it keeps coming up when i try to use the ± symbol

They have given you the equation of a quadratic, so the first thing you have to do is rearrange it, so that $x^{2}$ comes first.

So now we have the equation: 3 $x^{2}$  - 2x +1

*the quadratic formula uses a, b and c, but they are specific. a is always the coefficient of $x^{2}$ , b is always the coefficient of x and c is always the constant (number without an x)

So, i can find that, in this case:

a = 3 b= -2 c=1

Now substitute each of those values it into the original quadratic equation:

$\frac{-(-2) ± \sqrt{(-2)^{2} -4(3)(1)} }{2(3)}$

*again, ignore the weird capital A.

That's how you find the answer, which from what i can tell in the picture is C.

Hope that helped : )

6 0

3 years ago