How do you find the equation of the line in standard form that is perpendicular to the line y=3x+2 and passes through (-1, 5)please help!!!

Answers

Answer 1

Answer: $y=mx+b \perp y=3x+2 \n\Downarrow \hbox{the product of the slopes is -1} \nm * 3=-1 \nm=-(1)/(3) \n y=-(1)/(3)x+b \n \n(-1,5) \nx=-1 \ny=5 \n \Downarrow \n5=-(1)/(3) * (-1) + b \n5=(1)/(3)+b \n5-(1)/(3)=b \n(15)/(3)-(1)/(3)=b \nb=(14)/(3) \ny=-(1)/(3)x+(14)/(3) \n \ny=-(1)/(3)x+(14)/(3) \n(1)/(3)x+y=(14)/(3) \ \ \ |* 3 \n\boxed{x+3y=14}$

Answer 2

Answer: A perpendicular line has a slope that is the negative reciprocal of the line it's crossing. The negative reciprocal of 3 is -(1/3).

y=mx+b
5=-(1/3)(1)+b
5=(1/3)+b
4(2/3)=b

your final answer is:
y=(1/3)x+4(2/3)

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I need help ASAP pic below

(PLS HELP!!!!!!) Answer for a then b

Answers

Answer: Hey! Don't worry!! I gotchu'. :D

A. To find 1/5 of 15, we can multiply 15 by 1/5, which is the same as dividing 15 by 5. So, 1/5 of 15 is equal to 15 ÷ 5 = 3.

B. To find 4/5 of 35, we can multiply 35 by 4/5, which is the same as multiplying 35 by 4 and then dividing the result by 5. So, 4/5 of 35 is equal to (35 × 4) ÷ 5 = 28.

What are numbers that are not equal to 4.0

Answers

Answer:

4.1 and 4.2

Step-by-step explanation:

4.1 & 4.2

Well the person above was correct

Write an equation of the parabola in intercept form that passes through (−18, 72) with x-intercepts of −16 and −2.

Answers

y = a(x + 16)(x + 2)

72 = a(-18 + 16)(-18 + 2)

72 = a(-2)(-16)

72 = a(32)

$(72)/(32) = a$

$(9)/(4) = a$

Answer: y = $(9)/(4)$ (x + 16)(x + 2)

Can you help me with these please

Answers

Hello

1)
$(2x(x-3))/((x-1)(x+1))*(x(x-1))/((x-3))*(3(x+1))/(1)$
$=(6x(x+1))/(x+1)=6x$

2)
$(2(x-1)-(x+1)-(x+2))/((x-1)(x+1)(x+2))=(-5)/((x-1)(x+1)(x+2))$

Subtracting 3xy2 from 8xy2 gives the same result as the expression. A. 3xy^2 - 8xy^2 B. -7xy^2 - 2xy^2 C. 7xy^2 - 2xy^2