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Soloha48 [4]
3 years ago
14

90 POINTS MATH

Mathematics
2 answers:
satela [25.4K]3 years ago
6 0
<h3>Answer:</h3>

1.25 < y < 8.32124...

<h3>Step-by-step explanation:</h3>

The least y can be is the value that makes the angle in the smaller triangle be zero. (This condition will exist when the difference of side lengths is 4.)

... 4y -5 = 0

... y = 5/4 . . . . . add 5, divide by 4.

_____

The greatest y can be is the value obtained when the triangle is isosceles. For the larger triangle, that makes the side lengths (s) be ...

... 3/s = sin(43°/2)

... s = 3/sin(21.5°)

Then for the smaller triangle, that makes the angle be ...

... 4y -5 = arcsin(2/s) = arcsin(2/3·sin(21.5°))

... y = 1.25 + arcsin(2/3·sin(21.5°))/4

... y ≈ 8.32124

Amanda [17]3 years ago
3 0

Answer:

1.25 < y <= 8.32124

Step-by-step explanation:

We use the law of cosines for two triangles:

c^2 = a^2 + b^2 - 2 a b \cos(\gamma) \\ d^2 = a^2 + b^2 - 2 a b \cos(\delta)

This answer shows how to use the Reduce and Exists functions of Mathematica to solve either this problem, or the general problem of a pair of triangles with two sides of one triangle equal to two sides of the other triangle.  That answer (with c,\, d,\, \gamma,\, \text{and}\ \delta as independent variables which must be given ranges) has 11 cases, and would be a terrible waste of time to find by hand.

The law of cosines is used twice, with the same values for <em>a</em> and <em>b</em>, but different values for <em>c</em> and <em>γ</em> . Here I use the constants c = 6\,\ \gamma = 43°\, \ d = 4. The following equations and inequalities are supplied to Reduce, with an Extential quantifier specifying that Reduce should discover the range of values for \cos(\gamma).

\text{problem}=\exists _{\{a,b\}}\left(\begin{array}{ccc}36=a^2+b^2-2 a b \cos (43 {}^{\circ})\ \ \land\\16=a^2+b^2-2 a b \cos (\delta )\,\,\land\\ 0

This proves (since we used Reduce, not Solve, which is less reliable) that

  • a triangle exists that has angle 43°, two adjacent sides of length a and b and opposite side of length 6, and that
  • a second triangle exists with unknown angle, adjacent sides a and b equal to the corresponding sides of the first triangle, and opposite side length 4.
  • There is only one range of angles which satisfy the requirements.

\text{mincos}=\text{First}[\text{red}]\ \ \text{gives}\ \ \frac{1}{9} (4 \cos (43 {}^{\circ})+5)\\\text{maxcos}=\text{Last}[\text{red}]\ \ \text{gives}\ \ 1\\\\\text{Solve}[4 y-5=\delta,y ]\ \ \ \text{gives}\ \ \ \left\{\left\{y\to \frac{\delta }{4}+\frac{5}{4}\right\}\right\}\\ \\\\\ \frac{5}{4}

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Question 3 (Multiples and Factors] Three numbers are given below. Use prime factorisation to determine the HCF and LCM 1848 132
Ilia_Sergeevich [38]

Prime factorization involves rewriting numbers as products

The HCF and the LCM of 1848, 132 and 462​ are 66 and 1848 respectively

<h3>How to determine the HCF</h3>

The numbers are given as: 1848, 132 and 462

Using prime factorization, the numbers can be rewritten as:

1848 = 2^3 * 3 * 7 * 11

132 =  2^2 * 3 * 11

462 = 2 * 3 * 7 * 11

The HCF is the product of the highest factors

So, the HCF is:

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<h3>How to determine the LCM</h3>

In (a), we have:

1848 = 2^3 * 3 * 7 * 11

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462 = 2 * 3 * 7 * 11

So, the LCM is:

LCM = 2^3 * 3 * 7 * 11

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Read more about prime factorization at:

brainly.com/question/9523814

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2 years ago
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lawyer [7]

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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2. pemdas says that multiplication and division go before addition and subtraction, so ((6+(8)2÷4*1)/2)^4 becomes ((6+16÷4*1)/2)^4 ---> ((6+4)/2)^4

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