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2 years ago

For what values does c have no solutions?

Mathematics

1 answer:

defon2 years ago

3 0

Solve for c :

x⁴ + cx² + 100 = 0

cx² = -(x⁴ + 100)

c = - (x⁴ + 100)/x²

If x = 0, the right side is undefined and c would have no solution.

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2 years ago

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Shkiper50 [21]

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2 years ago

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

ANSWER:

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

SOLUTION:

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

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3 years ago

What is the answer

Reika [66]

Answer:

35 mi²

Step-by-step explanation:

Let's subdivide the figure, as shown.

The lower part is a rectangle whose area is (5 mi)(18 mi) = 90 mi².

The upper part is a trapezoid whose area is found by averaging the length and multiplying the result by the width (8 mi - 5 mi), or 3 mi.

Area of trapezoid:

12 mi + 18 mi

------------------------ = 15 mi Width of trapezoid = 3 mi

Thus, the area of the trapezoid is (3 mi)(15 mi) = 45 mi²

and the total area of the entire figure is

45 mi² + 90 mi² = 135 mi²

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3 years ago

What is 4 1/4 divied by 3 1/2

Ludmilka [50]

Answer:

1 3/14

Step-by-step explanation:

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2 years ago

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For what values ​​does c have no solutions?

For what values does c have no solutions?