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PtichkaEL [24]
2 years ago
12

How do you find a if given c and 0 (angle) in a right triangle where a is the opposite, c is the hypotenuse, b is adjacent and 0

is the angle in between c and b?
Mathematics
1 answer:
Yakvenalex [24]2 years ago
4 0
C can be found using pythagoras theorem. c2=a2+b2. Now, b is not given, but we know that cos(theta)=b/c=>b=c*cos(theta). Substituting b in the above relation, c2=a2+c2(cos(theta))^2=>c2=a2/(1-cos((theta))^2). c is the squareroot of c2. Hence c=sqrt(2/(1-cos((theta))^2))
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AysviL [449]

Answer:

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Step-by-step explanation:

6 0
2 years ago
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A deck of cards contains 6 black cards and 6 red cards. The first card drawn is not replaced before drawing the second card.
exis [7]

Answer : 5/22

A deck of cards contains 6 black cards and 6 red cards. The first card drawn is not replaced before drawing the second card.

Total number of outcome of cards = 12

Probability( any event)= favorable outcomes divide by total number of outcomes

Probability of selecting black card = \frac{6}{12} =\frac{1}{2}

The drawn card is not replaced . so 11 cards left . In that 11 cards, 5 black cards because one black card is already chosen

Probability of selecting another black card = \frac{5}{11}

Probability of selecting a black card followed by another black card =

= \frac{1}{2}*\frac{5}{11}

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4 0
3 years ago
1.) Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth .x squared minus 21 x equals n
Andrew [12]
Part 1
We are given x^2-21x=-4x. This can be rewritten as x^2-18x=0.
Therefore, a=1, b=-18, c=0.
Using the quadratic formula
     x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-18\right)\pm \sqrt{\left(-18\right)^2-4\left(1\right)\left(0\right)}}{2\left(1\right)}
     x=\frac{18\pm 18}{2}

The values of x are
     x_1=\frac{18-18}{2}=0
     x_2=\frac{18+18}{2}=18

Part 2
Since the values of y change drastically for every equal interval of x, the function cannot be linear. Therefore, the kind of function that best suits the given pairs is a quadratic function. 

Part 3.
The first equation is y=x^2+2.
The second equation is y=3x+20.

We have 
     x^2+2=3x+20
     x^2-3x-18=0
Factoring, we have 
     \left(x-6\right)\left(x+3\right)=0
Equating both factors to zero.
     x_1-6=0\rightarrow x_1=6
     x_2+3=0\rightarrow x_2=-3

When the value of x is 6, the value of y is 
     y=3\left(6\right)+20=38

When the value of x is -3, the value of y is 
     y=3\left(-3\right)+20=11

Therefore, the solutions are (6,38) or (-3,11)
7 0
2 years ago
You and your friends plan to attend the county fair this weekend. Admission to the fair is $5 and the cost per ride is $0.50. Ho
Luba_88 [7]
$20- $5= $15 for rides. You can go on 30 rides.
6 0
2 years ago
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3(9-8x-4x)+8(3x+4)=11<br> how do I work this out???
Arte-miy333 [17]
Remember PEMDAS is the order of operations. parenthesis,exponents,multiply,divide,add,subtract. so -8x-4x is -12x. now simplified the problem is 3(9-12x) + 8(3x+4)=11 . now I would distribute the numbers before the parentheses and it becomes 27-36x + 24x+32 =11 . now combine like terms. 59-12x =11 . subtract 59 on both sides. -12x=-48. divide -12 on both sides. x=4. :-)
7 0
2 years ago
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