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oksano4ka [1.4K]
3 years ago
12

If the sin of angle x is 4 over 5 and the triangle was dilated to be two times as big as the original, what would be the value o

f the sin of x for the dilated triangle?
Mathematics
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

Sin X will still be 4/5

Step-by-step explanation:

When we dilate, we are making bigger the dimensions of a shape

While we can dilate the measure of the sides, same cannot be said about the interior or exterior angles of the shape

this is because these angles maintain their original value

so while by dilating, we have the sides measuring 8 to 10

The ratio of the angle still will be the same

this is because 8/10 is still same 3/5

But, to the simplest term

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lina2011 [118]
Sorry I don’t now Gemini
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3 years ago
Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 9t + 9 cot(t/2), [π/4, 7π/4]
agasfer [191]

Answer:

the absolute maximum value is 89.96 and

the absolute minimum value is 23.173

Step-by-step explanation:

Here we have cotangent given by the following relation;

cot \theta =\frac{1 }{tan \theta} so that the expression becomes

f(t) = 9t +9/tan(t/2)

Therefore, to look for the point of local extremum, we differentiate, the expression as follows;

f'(t) = \frac{\mathrm{d} \left (9t +9/tan(t/2)  \right )}{\mathrm{d} t} = \frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9  \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2}

Equating to 0 and solving gives

\frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9  \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2} = 0

t=\frac{4\pi n_1 +\pi }{2} ; t = \frac{4\pi n_2 -\pi }{2}

Where n_i is an integer hence when n₁ = 0 and n₂ = 1 we have t = π/4 and t = 3π/2 respectively

Or we have by chain rule

f'(t) = 9 -(9/2)csc²(t/2)

Equating to zero gives

9 -(9/2)csc²(t/2) = 0

csc²(t/2)  = 2

csc(t/2) = ±√2

The solutions are, in quadrant 1, t/2 = π/4 such that t = π/2 or

in quadrant 2 we have t/2 = π - π/4 so that t = 3π/2

We then evaluate between the given closed interval to find the absolute maximum and absolute minimum as follows;

f(x) for x = π/4, π/2, 3π/2, 7π/2

f(π/4) = 9·π/4 +9/tan(π/8) = 28.7965

f(π/2) = 9·π/2 +9/tan(π/4) = 23.137

f(3π/2) = 9·3π/2 +9/tan(3·π/4) = 33.412

f(7π/2) = 9·7π/2 +9/tan(7π/4) = 89.96

Therefore the absolute maximum value = 89.96 and

the absolute minimum value = 23.173.

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4 years ago
Evin is building a tower out of paper cups. In each row (counting from the floor up) there are two less cups than the row below
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What does recursive mean.?
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3 years ago
I don’t understand :(
viktelen [127]

Answer:

The quotient when x^3-4x^2-8x+8 is divided by x+2 is x^2-6x+4 with no remainder, so x+2 is a factor of p(x).

Step-by-step explanation:

As the question suggests, there are two ways to solve this problem: long division (which can always be used to divide polynomials), and synthetic division (which can only be used when you are dividing by something of the form (x-a). In this case, we may use synthetic division, since x+2 is equal to x-(-2). I will use synthetic division here as it is slightly faster than long division.

The -2 on the left represents that we are dividing by x-(-2), and the rest of the numbers in the top row are the coefficients of p(x): 1, -4, -8, and 8.

The first step in synthetic division is to being down the leading coefficient of the polynomial: in this case, the 1 (as indicated in red). Now, we multiply the 1 by -2. We get -2, which we place directly below the -4.

Next, we add directly down the column, (-4 + (-2) is equal to -6), and this answer is placed in the box below. We can continue this process, getting the coefficients 1, -6, 4, and 0 in the bottom row.

This is the answer: the quotient is x^2-6x+4, and the remainder is zero (which indicates that x+2 is a factor of p(x)).

--

Edit: long division

We may also solve this problem using long division (see the second image). The first step is to look at the leading coefficients: since x \times x^2 is equal to x^3, the first term in the quotient will be x^2. Since x^2 \times (x+2) is equal to x^3+2x^2, we must now subtract that from x^3-4x^2-8x+8.

We repeat the same process, as shown in the image. Since we eventually get to zero, the remainder is zero, and the polynomial at the top (x^2-6x+4) is the quotient.

<em>I know this might be difficult to follow, so please comment if you have any questions.</em>

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3 years ago
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