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2 years ago

Write the equation of the line through (−2, −1) and perpendicular to 2y=3x−5.

2 answers:

5 0

Y=-1

Move all terms containing x to the left side of the equation.

Subtract 3x from both sides of the equation.
2x-3x=−5

Subtract 3x from 2x

−x=-5

Multiply each term in -x=−5 by −1

Multiply each term in −x=−5 by −1

(-x) ⋅ -1=(-5) ⋅-1

X=(-5) ⋅-1

multiple -5 by -1

X=5

Since x=5 is a vertical line, the slope is undefined.
Undefined

The equation of a perpendicular line to
x=5 must have a slope that is the negative reciprocal of the original slope.

M perpendicular = -1/∞
The negative reciprocal of ∞ is 0

M perpendicular is 0

Since the slope for the perpendicular line is 0 ,the line is perpendicular to the y-axis. The equation of this line is of the form
y=c when c is any real number. The equation of the line at point (-2 ,1)

Y=-1

I hope you understand this

REY [17]2 years ago

4 0

Answer:

Equation: y=-2/3x- $2\frac{2}{3}$

Step-by-step explanation:

y=mx +b

2y=3x-5

y=1.5x-2.5

y=-2/3x+b

(m is the negative reciprocal)

-2= 2/3+b

b=-2 $\frac{2}{3}$

Equation: y=-2/3x- $2\frac{2}{3}$

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Please answer this question I need this urgent Help meeeee

iren [92.7K]

Answer:

$\frac{s}{2t}$

Step-by-step explanation:

When you have exponents above a like term and they are being multiplied together, you add them.

For example:

$a^{x} *a^{y} = a^{x + y}$

So let's group like terms in the numerator:

$4r^{3} r^{-5} s^{-2} s^{-1}t$ We can add terms like in the example.

$4 r^{-2} s^{-3} t$

Let's rearrange the denominator.

$8r^{-2} s^{-4} t^{2}$

Now we have:

$\frac{4r^{-2} s^{-3}t}{8 r^{-2} s^{-4} t^{2} }$ Cancel like terms

4/8 = 1/2

$r^{-2} /r^{-2}$ = 1 So it cancels

$s^{-3} / s^{-4} = s^{-1}$ = s Since s is raised to the -1 it goes on top and becomes s.

$t / t^{2} = 1/t$

Now we combine everything back together:

$\frac{s}{2t}$

8 0

2 years ago

Matrix multiplication was used to encode a message using the given encoding matrix: [i 47 A=1 |-1 -3] The original message was c

nadya68 [22]

Answer:

$A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right]$ message SHOW_ME_THE_MONEY_

Step-by-step explanation:

The matrix

$A=\left[\begin{array}{cc}1&4\\-1&-3\end{array}\right]\rightarrow |A|=(1 \times -3)-(-1\times 4)=1\\\rightarrow A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] \\$

We can check that in fact A*A^⁻1=I_2 the identity matrix of size 2 x 2.

Now the message was divided in 1 x 2 matrices, then we have that the sequence given is the result of multiplying m by A, so to get m again we multiply now by A^⁻1. and we get the next table

Encoded message Decoded message message in letters by association

11 52 19 8 S H

-8 -9 15 23 O W

-13 -39 0 13 _ M

5 20 5 0 E _

12 56 20 8 T H

5 20 5 0 E _

-2 7 13 15 M O

9 41 14 5 N E

25 100 25 0 Y _

Then the message decoded is SHOW_ME_THE_MONEY_

3 0

2 years ago

X²-10x+25=0 quadratic equation by factoring

lisov135 [29]

Answer:

x = 5

Step-by-step explanation:

${x}^{2} - 10x + 25 = 0 \\ {x}^{2} - 5x - 5x + 25 = 0 \\ x(x - 5) - 5(x - 5) = 0 \\ (x - 5)(x - 5) = 0 \\ ( x- 5) = 0 \: or \: (x - 5) = 0 \\ x = 5 \: or \: x = 5 \\ x = 5$