Help please this is due at 12 !!!

Mathematics

1 answer:

Blababa [14]2 years ago

8 0

Answer:

Angle BC, because 70+20 = 90

Step-by-step explanation:

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What is the vertex of the graph y = x^2 + 6x - 5 ?

Dafna11 [192]

Given: <span>y = x^2 + 6x - 5. Then a = 1, b = 6 and c = -5.

The x-coord. of the vertex is given by x = -b / (2a), which here is x = -6 / (2*1) = -3.

Use the given formula </span><span>y = x^2 + 6x - 5 to find the value of y when x = -3:

y = (-3)^2 + 6(-3) - 5 = 9 - 18 - 5 = -14

Then the vertex is (-3, -14).</span>

6 0

3 years ago

Please help! Segment QR is tangent to circle P at point Q. What is the measure of angle PQR? A.) 37 degrees B.) 53 degrees C.) 9

Andrew [12]

Answer:

Correct option: C) 90 degrees

Step-by-step explanation:

When we have a tangent segment to a circle, the angle made with the radius of the circle and the tangent segment is always a right angle, that is, 90 degrees. In other words, the radius and the tangent segment are always perpendicular.

If QR is tangent to the circle P, and PQ is the radius of the circle, the segment PQ is perpendicular to the segment QR, so the angle PQR is equal 90 degrees.

Correct option: C)

6 0

3 years ago

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$