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Directions: For the following systems, determine whether or not the ordered triple (1, -1, -2) is a solution to the equations be

uranmaximum [27]

Put the coordinate and check each

#1

$\\ \rm\Rrightarrow 3x-3y-z=2$

$\\ \rm\Rrightarrow 3(1)-3(-1)-(-2)=2$

$\\ \rm\Rrightarrow 3+3+2=2$

$\\ \rm\Rrightarrow 6+2=2$

$\\ \rm\Rrightarrow 8\neq 2$

No

#2

$\\ \rm\Rrightarrow 2(1)-3(-1)-2=-3$

$\\ \rm\Rrightarrow 2+3-2=-3$

$\\ \rm\Rrightarrow 3\neq-3$

Unless you did typing mistake on right hand side it's

No

#3

$\\ \rm\Rrightarrow 1-1-2=-2$

$\\ \rm\Rrightarrow -2=-2$

Yes

#4

$\\ \rm\Rrightarrow 1+2(-1)-(-2)=0$

$\\ \rm\Rrightarrow 1-2+2=0$

$\\ \rm\Rrightarrow 1\neq0$

No

#5

$\\ \rm\Rrightarrow 5(1)+1-5(-2)=16$

$\\ \rm\Rrightarrow5+1+10=16$

$\\ \rm\Rrightarrow 16=16$

Yes

#6

$\\ \rm\Rrightarrow 4(1)+5(-1)-6(-2)=-13$

$\\ \rm\Rrightarrow 4-5+12=-13$

$\\ \rm\Rrightarrow -1+12=-13$

$\\ \rm\Rrightarrow 11\neq -13$

No

8 0

2 years ago

The length of the line segment containing the points (1,7) and (5,5)

ArbitrLikvidat [17]

Answer:

True

Step-by-step explanation:

Let A denotes the point (1,7)

Let B denotes the point (5,5)

We are supposed to find The length of the line segment containing the points

Formula : $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(x_1,y_1)=(1,7)\\(x_2,y_2)=(5,5)\\ d = \sqrt{(5-1)^2+(5-7)^2}\\ d = \sqrt{4^2+(-2)^2}\\ d = \sqrt{4^2+(-2)^2}\\d=4.47$

So,The length of the line segment containing the points (1,7) and (5,5) is 4.47 units is true.

Hence Option A is true

4 0

3 years ago

Multiply. Write your answer as a fraction or as a whole or mixed number.

I am Lyosha [343]

Answer:
4 1/6

5 x 5 = 25
6 x 1 = 6
Simplify
25/6 = 4 1/6

4 0

3 years ago

Simplify each expression by combining like terms. 1) 3y+4x-y+2x 2) 3y+2+3x-y+5x

SVEN [57.7K]

Answer:

Step-by-step explanation:

1) 2y + 6x

2) 2y + 8x +2

8 0

3 years ago

Pls help this is pretty urgent

KIM [24]

Answer:

(a) 0

(b) f(x) = g(x)

(c) See below.

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{x^2+2x+1}{x^2-1}$

Part (a)

Factor the numerator and denominator of the given rational function:

$\begin{aligned} \implies f(x) & = \dfrac{x^2+2x+1}{x^2-1} \\\\& = \dfrac{(x+1)^2}{(x+1)(x-1)}\\\\& = \dfrac{x+1}{x-1}\end{aligned}$

Substitute x = -1 to find the limit:

$\displaystyle \lim_{x \to -1}f(x)=\dfrac{-1+1}{-1-1}=\dfrac{0}{-2}=0$

Therefore:

$\displaystyle \lim_{x \to -1}f(x)=0$

Part (b)

From part (a), we can see that the simplified function f(x) is the same as the given function g(x). Therefore, f(x) = g(x).

Part (c)

As x = 1 is approached from the right side of 1, the numerator of the function is positive and approaches 2 whilst the denominator of the function is positive and gets smaller and smaller (approaching zero). Therefore, the quotient approaches infinity.

$\displaystyle \lim_{x \to 1^+} f(x)=\dfrac{\to 2^+}{\to 0^+}=\infty$

5 0

1 year ago

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