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vichka [17]
3 years ago
5

¿Cual es el mínimo común múltiplo de 20 14 y 17? Gracias ♥ ♥

Mathematics
1 answer:
Scorpion4ik [409]3 years ago
8 0

El mínimo común múltiplo de 20 14 y 17 es <u>1</u>.

17 es un número primo, y su únicos múltiplos son 17 y 1.

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Add:(3x2 - 5x + 6) + (9 - 8x - 4x2)​
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Which of the following are the coordinates of the vertex of y = −x2 + 2x + 1? (2, 1) (1, –2) (1, 2) (–2, 1)
IceJOKER [234]

Answer:

C

Step-by-step explanation:

We want to determine the vertex of the quadratic equation:

\displaystyle y = -x^2 + 2x + 1

Recall that the vertex is given by the formulas:

\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)

In this case, <em>a</em> = -1, <em>b</em> = 2, and <em>c</em> = 1.

Determine the <em>x-</em>coordinate of the vertex:

\displaystyle x = -\frac{(2)}{2(-1)} = 1

To determine the <em>y-</em>coordinate, evaluate the function at <em>x</em> = 1:

\displaystyle \begin{aligned} y(1)&= -(1)^2 + 2(1) + 1 \\ &= 2\end{aligned}

In conclusion, the vertex of the quadratic equation is (1, 2).

Hence, our answer is C.

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2 years ago
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Read 2 more answers
A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

               =\frac{24\times (-13)}{\sqrt{676}}

               =\frac{24\times (-13)}{26}

               = -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0
3 years ago
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