MATHEMATICS HIGH SCHOOL

I will give you brainliest!!! Solve the system of equations by graphing both equations with a pencil and paper y=x-3 y=-x+1

Answers

Answer 1

Answer:

Answer:

A. (2,-1) i thought this is hard but its kind of easy actually

Step-by-step explanation:

so you just put the given options into the xs and ys.. so lets say we have (2,-1) 2=x -1=y

now put them into the equations.. y=x-3

-1=2-3( if this is true than this is the right answer)

y= -x +1

-1= -2 +1 (this is also correct)

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The monthly electric bills in acity have a mean of $86 and a
standard deviation of $18. Find
the z-scores that correspond
to a utility bill of $160.

Answers

Answer:

Step-by-step explanation:

Find the value of the expression
2r x s
for r = 2 and s = 4.

Answers

Answer:

Step-by-step explanation:

2rs

Let r=2 and s=4

2*2*4

4*4

Is -2/7 a rational or irrational number

Answers

Answer:

Rational

Step-by-step explanation:

All fractions are rational.

Explain the derivation behind the derivative of sin(x) i.e. prove f'(sin(x)) = cos(x)How about cos(x) and tan(x)?

Answers

1.

$f'(\sin x) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\sin(x+h) - \sin(x))/(h) = \n \n = \lim_(h \to 0) (2 \sin( (x+h - x)/(2)) \cdot \cos( (x+h+x)/(2)) )/(h) = \lim_(h \to 0) (2 \sin( (h)/(2)) \cos( (2x+h)/(2) ) )/(h) = \n \n = \lim_(h \to 0) [ (\sin( (h)/(2)) )/( (h)/(2) ) \cdot \cos ((2x+h)/(2)) ] = \lim_(h \to 0) [1 \cdot \cos( (2x+h)/(2) ) ] =$

$= \cos( (2x)/(2)) = \boxed{\cos x}$

2.

$f'(\cos x) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\cos(x+h) - \cos(x))/(h) = \n \n = \lim_(h \to 0) (-2 \sin ( (x+h+x)/(2)) \cdot \sin ( (x+h-x)/(2)) )/(h) = \lim_(h \to 0) (-2 \sin ( (2x+h)/(2)) \cdot \sin ( (h)/(2)) )/(h) = \n \n = \lim_(h \to 0) (-2 \sin ( (2x+h)/(2)) )/(2) \cdot (sin( (h)/(2)) )/( (h)/(2) ) = \lim_(h \to 0) -\sin( (2x+h)/(2)) \cdot 1 =$

$= -\sin( (2x)/(2)) = \boxed{\sin x }$

3.

$f'(\tan) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\tan(x+h) - \tan(x))/(h) = \n \n = \lim_(h \to 0) ( (\sin(x+h-x))/(\cos(x+h) \cdot \cos(x)) )/(h) = \lim_(h \to 0) ( (\sin(h))/( (\cos(x+h-x) + \cos(x+h+x))/(2) ) )/(h) =$

$= \lim_(h \to 0) ( (\sin(h))/(\cos(h) + \cos(2x+h)) )/( (1)/(2)h ) = \lim_(h \to 0) (\sin(h))/( (1)/(2)h \cdot [\cos(h) + \cos(2x+h)] ) = \n \n = \lim_(h \to 0) (\sin(h))/(h) \cdot (1)/( (1)/(2) \cdot (\cos(h) + cos(2x+h) ) = 1 \cdot (1)/( (1)/(2) \cdot (1+ cos(2x) ) = (2)/(1 + 2 \cos^(2) - 1 ) = \n \n = (2)/(2 \cos^(2) x) = \boxed{ (1)/(\cos^(2)x) }$

4.

$f'(\cot) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\cot(x+h) - \cot(x))/(h) = \n \n = \lim_(h \to 0) ( (\sin(x - x - h))/(\sin (x+h) \cdot \sin (h)) )/(h) = \lim_(h \to 0) ( (\sin(-h) )/( (\cos(x+h-x) - \cos(x+h+x))/(2) ) )/(h) =$

$= \lim_(h \to 0) ( (-\sin(h))/(\cos(h) - \cos(2x+h)) )/( (1)/(2)h ) = \lim_(h \to 0) ( - \sin(h))/( (1)/(2)h \cdot [\cos(h) - \cos(2x+h)] ) = \n \n = \lim_(h \to 0) (- \sin (h))/(h) \cdot (1)/( (1)/(2) \cdot [\cos(h) - \cos(2x+h)] ) = -1 \cdot (2)/(1 - cos(2x)) = \n \n = - (2)/(1 -1 + 2 \sin^(2)x) = - (2)/(2 \sin^(2) x) = \boxed{- (1)/(\sin^(2) x) }$

I posted an image instead.

Marlon asks a friend to think of a number from 5 to 11. What is the probability that Marlon’s friend will think of the number 9?

Answers

Answer:

$P=(1)/(7)$

Step-by-step explanation:

we know that

The probability of an event is the ratio of the size of the event space to the size of the sample space.

The size of the sample space is the total number of possible outcomes

The event space is the number of outcomes in the event you are interested in.

Let

x---------> size of the event space

y-------> size of the sample space

$P=(x)/(y)$

In this problem we have

$x=1$ (because is only one number to think)

$y=7$ (there are $7$ numbers between $5$ and $11$ )

substitute

$P=(1)/(7)$

There are 7 numbers between 5 and 11 including 5 and 11. This means there is a one in seven chance of any number. The probability of o 9 is 1/7.

Rewrite the function y = x2 - 6x + 14

Answers

y = x² - 6x + 14

rewrite in vertex form:

y - 14 = x² - 6x

perfect square trinomial.
x² ⇒ x * x
-6x ⇒ -6x / 2 = -3x
-3² = 9

(x - 3)² ⇒ (x-3)(x-3) ⇒ x(x-3) -3(x-3) = x² - 3x - 3x + 9 = x² - 6x + 9

y - 14 + 9 = (x-3)²
y -5 = (x-3)²
y = (x-3)² + 5