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4 years ago

12

Mathematics

1 answer:

weeeeeb [17]4 years ago

8 0

Answer:

2x + y = -4

Step-by-step explanation:

standard form of equation of straight line is

ax+by = c

that is terms containing x and y should be on LHS and constant term should be on RHS

______________________________________________

Given equation

y + 1 = - 2x - 3

lets bring -2x on LHS ,

add 2x on lHS and RHS

y + 1 + 2x = - 2x - 3 + 2x

=> y + 1 + 2x = -3

on lHS, 1 is there which constant term lets bring it on RHS

subtract 1 from both sides

y + 1 + 2x - 1= -3 -1

y + 2x = -4

rearranging it

2x + y = -4 (Answer)

You might be interested in

Zack watered his garden with 1 3/8

victus00 [196]

Answer:

The correct answer is that Zack used 2 7/8 gallons to water his garden on the fifth week.

Step-by-step explanation:

Let's find out the pattern Zack is using to water his garden, this way:

First week = 1 3/8 gallons of water

Second week = 1 3/4 gallons of water = 1 3/8 + 3/8 = 1 6/8 = 1 3/4

Third week = 2 1/8 gallons of water = 1 3/4 + 3/8 = 1 6/8 + 3/8 = 2 1/8

The constant of adding every week is 3/8 gallons, therefore,

Fourth week = 2 1/8 + 3/8 = 2 4/8 = 2 1/2 gallons of water

Fifth week = 2 1/2 + 3/8 = 2 4/8 + 3/8 = 2 7/8 gallons of water

The correct answer is that Zack used 2 7/8 gallons to water his garden on the fifth week.

3 0

3 years ago

Show work and explain with formulas:

Marina86 [1]

17 Answer: 205

Step-by-step explanation:

$\{1\dfrac{2}{3}+1\dfrac{5}{6}+2+...+8\dfrac{1}{3}\}\implies a_1=1\dfrac{2}{3},\ d=\dfrac{1}{6}\\\\\\a_n=a_1+d(n-1)\qquad solve\ for\ n\\\\8\dfrac{1}{3}=1\dfrac{2}{3}+\dfrac{1}{6}(n-1)\\\\\\\dfrac{25}{3}=\dfrac{5}{3}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{50}{6}=\dfrac{10}{6}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{41}{6}=\dfrac{1}{6}n\\\\\\\dfrac{41}{6}\cdot 6=n\\\\41=n$

$\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{41}=\dfrac{1\dfrac{2}{3}+8\dfrac{1}{3}}{2}\cdot 41\\\\\\.\quad =\dfrac{10}{2}\cdot 41\\\\\\.\quad =5\cdot 41\\\\.\quad =\large\boxed{205}$

18 Answer: 1968

Step-by-step explanation:

$a_1=-6,\ d=2,\ n=48, \quad \text{solve for }a_{48}\\\\a_{n}=a_1+d(n-1)\\\\a_{48}=-6+2(48-1)\\\\.\quad =-6+2(47)\\\\.\quad =-6+94\\\\.\quad =88$

$\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{48}=\dfrac{-6+88}{2}\cdot 48\\\\\\.\quad =\dfrac{82}{2}\cdot 48\\\\\\.\quad =41\cdot 48\\\\.\quad =\large\boxed{1968}$

19 Answer: -116

Step-by-step explanation:

$\{3, -2, -7, ...\}\\a_1=3,\ d=-5,\ n=8, \quad \text{solve for }a_{8}\\\\a_{n}=a_1+d(n-1)\\\\a_{8}=3-5(8-1)\\\\.\quad =3-5(7)\\\\.\quad =3-35\\\\.\quad =-32\\\\\text{Now use the sum formula:}\\\\S_8=\dfrac{a_1+a_8}{2}\cdot 8\\\\\\S_{8}=\dfrac{3-32}{2}\cdot 8\\\\\\.\quad =\dfrac{-29}{2}\cdot 8\\\\\\.\quad =-29\cdot 4\\\\.\quad =\large\boxed{-116}$