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Eva8 [605]
3 years ago
5

Pythagoras Theorem and Quadratic Equations Solve for x: a= (x-2) b= 9 c= (x+1)

Mathematics
2 answers:
UkoKoshka [18]3 years ago
6 0

Answer:

x = -13

Step-by-step explanation:

using Pythagoras theorem which says that the square of the hyponeneus side of a right angled triangle is equal to the sum of square of the two opposite sides.

mathematically,

|H|² = |OPP|² + |ADJ|²

H= (x-2) , opp= 9 adj=(x+1)

(x-2)² = (9)² + (x+1)²

x² -4x + 4 = 81 + x²+2x+1

collecting the like terms

x²- x² - 4x-2x = 81 +1 - 4

- 6x = 89 -4

-6x = 78

divide both sides by the coefficient of x which is -6

-6x/-6 = 78/-6

x = -13

Alex_Xolod [135]3 years ago
3 0

Answer:

Step-by-step explanation:

Pythagoras theorem

a^2=b^2+c^2

(x-2)^2=9^2+(x+1)^2

x^2-4x+4=81+x^2+2x+1

Collect like terms

-4x-2x=82-4

-6x=78

x=-13

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Which function f (x) , graphed below, or g (x) , whose equation is g (x) = 3 cos 1/4 (x + x/3) + 2, has the largest maximum and
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Answer:

g(x), and the maximum is 5

Step-by-step explanation:

for given function f(x), the maximum can be seen from the shown graph i.e. 2

But for the function g(x), maximum needs to be calculated.

Given function :

g (x) = 3 cos 1/4 (x + x/3) + 2

let x=0 (as cosine is a periodic function and has maximum value of 1 at 0 angle)

g(x)= 3 cos1/4(0 + 0) +2

    = 3cos0 +2

     = 3(1) +2

     = 3 +2

     = 5 !

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Reduce fraction 27/36 lowest term
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When an automobile is stopped with a roving safety patrol,each tire is checked for tire wear, and each headlight is checkedto se
ryzh [129]

Answer:

a) Joint ptobability distribution

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

P(x,y)=P_x(x)*P_y(y)

Then, the joint probabilities are:

X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025

X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06

X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01

X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56

We can calculate P(X<= 1) * P(Y<=1)

P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56

Both calculations give the same result.

c) Probability of no violations

P(X+Y=0)=P(0,0)=0.3

d) P(X + Y <= 1)

P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53

5 0
3 years ago
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