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sertanlavr [38]
3 years ago
12

Can anyone help?

Mathematics
1 answer:
SVEN [57.7K]3 years ago
4 0

For the equation (2k + 1)x² + 2x = 10x - 6 to have two real and equal roots, the value of k = 5/6.

Since the equation is (2k + 1)x² + 2x = 10x - 6, we collect subtract 10x from both sides and add 6 to both sides.

So, we have (2k + 1)x² + 2x - 10x + 6 = 10x - 6 - 10x + 6

(2k + 1)x² - 8x + 6 = 0

For the equation, (2k + 1)x² + 2x = 10x - 6 to have two real and equal roots, this new equation (2k + 1)x² - 8x + 6 = 0 must also have two real and equal roots.

For the equation to have two real and equal roots, its discriminant, D = 0.

D = b² - 4ac where b = -8, a = 2k + 1 and c = 6.

So, D =  b² - 4ac

D =  (-8)² - 4 × (2k + 1) × 6 = 0

64 - 24(2k + 1) = 0

Dividing through by 8, we have

8 - 3(2k + 1) = 0

Expanding the bracket, we have

8 - 6k - 3 = 0

Collecting like terms, we have

-6k + 5 = 0

Subtracting 5 from both sides, we have

-6k = -5

Dividing through by -6, we have

k = -5/-6

k = 5/6

So, for the equation (2k + 1)x² + 2x = 10x - 6 to have two real and equal roots, the value of k = 5/6.

Learn more about quadratic equations here:

brainly.com/question/18162688

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Find the length of side e to the nearest tenth in the triangle below.
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Answer:12.1

Step-by-step explanation:

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Angle D=180-134

Angle D=46

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3 years ago
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a_sh-v [17]

The range of the equation is y>2

Explanation:

The given equation is y=2(4)^{x+3}+2

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

y=2 \cdot 4^{x+3}+2

This can be written as y=2^{1+2(x+3)}+2

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

x=2^{1+2(y+3)}+2

Solving for y, we get;

x-2=2^{1+2(y+3)}

Applying the log rule, if f(x) = g(x) then \ln (f(x))=\ln (g(x)), then, we get;

\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)

Simplifying, we get;

(1+2(y+3)) \ln (2)=\ln (x-2)

Dividing both sides by \ln (2), we have;

2 y+7=\frac{\ln (x-2)}{\ln (2)}

Subtracting 7 from both sides of the equation, we have;

2 y=\frac{\ln (x-2)}{\ln (2)}-7

Dividing both sides by 2, we get;

y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}

Let us find the positive values for logs.

Thus, we have,;

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The function domain is x>2

By combining the intervals, the range becomes y>2

Hence, the range of the equation is y>2

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