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3 years ago

Tell whether the angles are complementary or supplementary. then find the value of x

Mathematics

2 answers:

chubhunter [2.5K]3 years ago

7 0

Hello,

In Q5. Being a straight line the angles are supplementary to each other.

And x = 70°

In Q6.

The angles sum up to 90° , so 25x + 30° = 90

25x = 60°

x= 60°/25 = 2.4°

Hope this helps.

ratelena [41]3 years ago

6 0

Answer:

see explanation

Step-by-step explanation:

x and 110 form a straight angle and are supplementary, thus

x + 110 = 180 ( subtract 110 from both sides )

x = 70

25x and 30 form a right angle and are complementary, thus

25x + 30 = 90 ( subtract 30 from both sides )

25x = 60 ( divide both sides by 25 )

x = $\frac{60}{25}$ = $\frac{12}{5}$

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3 years ago

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Answer:

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Step-by-step explanation:

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Pani-rosa [81]

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11. Recordemos que entre más negativo es un número, menor es.

a) - 6 > - 10

ya que - 10 es más negativo, por consiguiente es menor, esto llega a confundir ya que tenemos la noción de que si vemos una cifra alta, por consiguiente es mayor, esto pasa en los positivos.

b) 12 > - 16

c) - 7 < 0

d) 35 > - 80

e) -37 < - 24

f) - 40 < 15

g) - 18 < 10

h) - 17 < 0

i) |-21| = |21|

El valor absoluto de un número negativo es su positivo, por lo que también puede ser:

21 = 21.

j) |176| > 0, o también 176 > 0

k) |-479| > |478|, o también 479 > 478

l) |-1375| = |1375|, o también 1375 = 1375

12. De menor a mayor.

a) - 8, - 7, - 4, 0, 3

b) - 16, - 12, - 7, 8, 9

c) - 15, - 13, 0, 9, 11

d) - 47, - 43, - 31, 14, 29

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f) - 26, - 21, - 10, 17, 26

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Salu2

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3 years ago

Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis

stealth61 [152]

Complete question is;

Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola. (Round your answers to the nearest hundredth.) y = 6 - x²

Answer:

height = 4 and base = 2√2

Step-by-step explanation:

Area of a rectangle is given by;

A = base(b) × height(h)

To start off, we will first have to consider half of the rectangle.

Due to the fact that it is bounded by a parabola which is symmetric over the y-axis, it means that the rectangle will have the same area on the right and left hand sides.

So if we maximize the area of the first quadrant, it means that we are also maximizing the area of the entire rectangle.

Now, since we are dealing with the base and other 2 vertices on the x-axis, it means that the base(b) of this half rectangle will be x.

The height of this rectangle is the y coordinate of the corner that is directly above the x coordinate, which is also on the rectangle.

Therefore, the coordinates of the upper corner will be (x, y), or (x, 6 - x²). Thus our half base is x and full height is 6 - x². Area will be;

A = x(6 - x²)

A = 6x - x³

We will maximize this area by finding the derivative and equating to zero.

Thus;

A' = 6 - 3x²

At A' = 0,we have;

6 - 3x² = 0

3x² = 6

x² = 6/3

x² = 2

x = √2

Thus, from y = 6 - x², we have;

y = 6 - (√2)²

y = 6 - 2

y = 4

Since x is half base, it means our full width is 2x = 2 × √2 = 2√2

So, height = 4 and base = 2√2

8 0

3 years ago

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information: $P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000$ , where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

$\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2$

Equating it to zero we get,

$x^2 + 36x - 6400 = 0$

We use the quadratic formula to find the values of x:

$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ , where a, b and c are coefficients of $x^2, x^1 , x^0$ respectively.

Putting these value we get x = -100, 64

Now, again differentiating

$\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x$

At x = 64, $\displaystyle\frac{d^2(P(x))}{dx^2} < 0$

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0

3 years ago