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3 years ago

What is the sqrt{-38} in the simplest form

2 answers:

5 0

I’m assuming you know complex numbers,

That’s the principle square root, so it’s going to be positive.

Sqrt(-38) = sqrt(-1)*sqrt(2)*sqrt(19)

Sqrt(-38) = isqrt(38)

Where i^2 = -1

You could break up 38 into 2 times 19, but that makes it messier.

mixas84 [53]3 years ago

3 0

Answer: 6.164 √38

Step-by-step explanation: hope this is what your looking for :)

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11 ounces is to one cup as<br> ounces is to 10 cups.

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Answer:

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3 years ago

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If θ is an angle in standard position and its terminal side passes through the point (-5,-6), find the exact value of

Mariana [72]

Answer:

$\sec \theta = -\frac{\sqrt{61}}{5}$

Step-by-step explanation:

The secant can be found by the following trigonometric relation:

$\sec \theta = \frac{1}{\cos \theta}$

$\sec \theta = \frac{1}{\frac{x}{r} }$

$\sec \theta = \frac{r}{x}$

$\sec \theta = \frac{\sqrt{x^{2}+y^{2}}}{x}$

$\sec \theta = \frac{\sqrt{(-5)^{2}+(-6)^{2}}}{(-5)}$

$\sec \theta = -\frac{\sqrt{61}}{5}$

6 0

3 years ago

The standard form of a quadratic equation is given as y=ax^2+bx+c and vertex form is given as y=a(x−h)^2+k. Write a formula that

Tju [1.3M]

Answers:

$h = -\frac{b}{2a}\\\\k = \frac{-b^2+4ac}{4a}\\\\$

where 'a' cannot be zero.

=========================================================

Explanation:

The vertex is (h,k)

The x coordinate of the vertex is h which is found through this formula

$x = -\frac{b}{2a}$

For example, if we had the quadratic $y = 3x^2-6x+5$ , then we'll plug in a = 3 and b = -6 to get: $h = -\frac{b}{2a} = -\frac{-6}{2*3} = 1$

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To find the value of k, we plug that h value into the original standard form of the quadratic and simplify.

$y = ax^2+bx+c\\\\k = ah^2+bh+c\\\\k = a\left(\frac{-b}{2a}\right)^2+b\left(\frac{-b}{2a}\right)+c\\\\k = a*\frac{b^2}{4a^2}+\frac{-b^2}{2a}+c\\\\k = \frac{b^2}{4a}+\frac{-b^2}{2a}+c\\\\$

$k = \frac{b^2}{4a}+\frac{-b^2}{2a}*\frac{2}{2}+c*\frac{4a}{4a}\\\\k = \frac{b^2}{4a}+\frac{-2b^2}{4a}+\frac{4ac}{4a}\\\\k = \frac{b^2-2b^2+4ac}{4a}\\\\k = \frac{-b^2+4ac}{4a}\\\\$

It's interesting how we end up with the numerator of $-b^2+4ac$ which is similar to $b^2-4ac$ found under the square root in the quadratic formula. There are other ways to express that formula above. We need $a \ne 0$ to avoid dividing by zero. The values of b and c are allowed to be zero.