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3 years ago

For every quad, there are 4 wheels. Complete the

Mathematics

1 answer:

PIT_PIT [208]3 years ago

7 0

The one next to 3 is 12, the one next to 7 is 28, the one next to 12 is 48. All you have to do is multiply the quads by four and you get the number of wheels.

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What is the graph of f(x)=x^2-2x+3

Neko [114]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$f(x)=x^{2}-2x+3$

This is the equation of a vertical parabola open upwards

The vertex is a minimum

The vertex is the point (1,2)

The y-intercept is the point (0,3)

The function does not have x-intercepts

see the attached figure

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3 years ago

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Step by step explanation

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3 years ago

Urgent, please i will give brainliest

Ne4ueva [31]

Answer:

1. $\frac{\sqrt{13} }{3}$

2. 3.2362

3. $\frac{60}{11}$

Step-by-step explanation:

Cot is the trigonometric ratio defined by "adjacent" over "opposite". So, adjacent = 2 and opposite = 3.

By pythagorean theorem, we have the "hypotenuse" as $\sqrt{2^2+3^2}=\sqrt{13}$

Csc is defined as the trig ratio "hypotenuse" over "opposite". We know the sides, so Csc $\theta$ = $\frac{\sqrt{13} }{3}$

First answer choice is right.

By definition, Csc $\theta$ is the inverse of Sine $\theta$ . If given the value of sin theta, to find value of csc theta, we take the reciprocal of it. Hence:

$Csc\theta=\frac{1}{0.3090}\\=3.2362$

Third answer choice is right.

By definition tan and cot are inverse of each other. So the value of tan is the reciprocal of the value of cot. We can simply "flip" the value of tan theta to get the value of cot theta. Hence:

$Cot\theta=\frac{60}{11}$

Third answer is right.

3 0

3 years ago

Read 2 more answers

A piece of wire 11 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

forsale [732]

Answer:

11 meters

Step-by-step explanation:

First, we can say that the square has a side length of x. The perimeter of the square is 4x, and that is how much wire goes into the square. To maximize the area, we should use all the wire possible, so the remaining wire goes into the triangle, or (11-4x).

The area of the square is x², and the area of an equilateral triangle with side length a is (√3/4)a². Next, 11-4x is equal to the perimeter of the triangle, and since it is equilateral, each side has (11-4x)/3 length. Plugging that in for a, we get the area of the equilateral triangle is

(√3/4)((11-4x)/3)²

= (√3/4)(11/3 - 4x/3)²

= (√3/4)(121/9 - 88x/9 + 16x²/9)

= (16√3/36)x² - (88√3/36)x + (121√3/36)

The total area is then

(16√3/36)x² - (88√3/36)x + (121√3/36) + x²

= (16√3/36 + 1)x² - (88√3/36)x + (121√3/36)

Because the coefficient for x² is positive, the parabola would open up and the derivative of the parabola would be the local minimum. Therefore, to find the maximum area, we need to go to the absolute minimum/maximum points of x (x=0 or x=2.75)

When x=0, each side of the triangle is 11/3 meters long and its area is

(√3/4)a² ≈ 5.82

When x=2.75, each side of the square is 2.75 meters long and its area is

2.75² = 7.5625

Therefore, a maximum is reached when x=2.75, or the wire used for the square is 2.75 * 4 = 11 meters

3 0

3 years ago