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Evgesh-ka [11]
3 years ago
15

I dont really know how to do soh cah toah. please help me find x.

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
4 0
First notice that the triangle with sides x,y,a and the triangle with sides z,a+b,x are similar. This is true because the angle between sides x,y in the smaller triangle is clearly 30^\circ, while the angle between sides y,z in the larger triangle is clearly 60^\circ. So the triangles are similar with sides x,y,a corresponding to a+b,z,x, respectively.

Now both triangles are 30^\circ-60^\circ-90^\circ, which means there's a convenient ratio between its sides. If the length of the shortest leg is \ell_1, then the length of the longer leg is \ell_2=\sqrt3\ell_1 and the hypotenuse has length \ell_3=\sqrt{{\ell_1}^2+{\ell_2}^2}=2\ell_1.

Since x is the shortest leg in the larger triangle, it follows that a+b=2x, so a+b=15=2x\implies x=\dfrac{15}2
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Find the vertices of the following functions.<br> Y=4(x-2)(x-3)<br><br> Y=-4(x+2)^2 -5
Natalija [7]
1) 5/2 , -1
2) -2,-5
8 0
3 years ago
A survey of 2,000 doctors showed that an average of 3 out of 5 doctors use brand X aspirin. How many doctors use brand X aspirin
swat32
You would use cross multiplication:

X/2,000 = 3/5
Multiply: 2,000*3= 6,000
Divide: 6,000/5
Answer: 1,200 doctors use brand x
8 0
3 years ago
5. Let A = (x, y), B = {1,2). Find the Cartesian products of A and B: A x B? (Hint: the result will be a set of pairs (a, b) whe
hichkok12 [17]

Answer: A x B = {(x,1), (x,2), (y,1), (y,2)}

Step-by-step explanation:

The Cartesian product of any two sets M and N is the set of all possible ordered pairs such that the elements of M are first values and the elements of N are the second values.

The Cartesian product of sets M and N is denoted by M × N.

For Example : M = {x,y} and N={a,b}

Then , M × N ={(x,a), (x,b), (y,a), (y,b)}

Given : Let A = {x, y}, B = {1,2}

Then , the Cartesian products of A and B will be :

A x B = {(x,1), (x,2), (y,1), (y,2)}

Hence, the Cartesian products of A and B = A x B = {(x,1), (x,2), (y,1), (y,2)}

8 0
3 years ago
There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student
Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0
3 years ago
A cell phone company has three different production sites.
dlinn [17]

Answer:

3/11

Step-by-step explanation:

given that a cell phone company has three different production sites

Site name                    1                              2                     3               Total

Production                  60%                         30%                10%             100%

Recalled                        5%                           7%                   9%                  

Prob for recall            0.003                       0.021             0.009         0.033

Total probability for recall = 0.033

Prob for recall and came from site 3 = 0.009

So required probability

= If a randomly selected cell phone has been recalled, what is the probability that it came from Site 3

=\frac{0.009}{0.033} \\=\frac{3}{11}

4 0
3 years ago
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