1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ki77a [65]
3 years ago
11

1. Change each of the following equations to the form

Mathematics
1 answer:
Evgen [1.6K]3 years ago
5 0

Answer:

(a) -6x- y = 3

        6x + y = -3\\      y = -6x -3

(b) 3x-4y =12

       3x - 4y = 12\\-4y = -3x +12\\4y = 3x -12\\\\y = \frac{3}{4}x -3

(c)-2x + 5y = 10

       -2x + 5y = 10\\5y = 2x + 10\\y = \frac{2}{5}x + 2

(d) x + 2y = 4

        x+2y=4\\2y = -x + 4\\y = \frac{-1}{2} x +2

(e) 8x- 4y = 3

        8x - 4y =3\\-4y = -8x + 3\\4y = 8x -3\\y = 2x -\frac{3}{4}

(f) 4x + 5y = -6

         4x + 5y = -6\\5y = -4x - 6\\y = \frac{-4}{5} x -\frac{6}{5}

You might be interested in
Select the two values of x that are roots of this equation.<br> 2x+11x+15= 0
Amiraneli [1.4K]

Answer:

The roots of the equation are x=-3 and x=-2.5

Step-by-step explanation:

<u><em>The correct quadratic equation is</em></u>

2x^2+11x+15=0

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

2x^{2} +11x+15=0  

so

a=2\\b=11\\c=15

substitute in the formula

x=\frac{-11(+/-)\sqrt{11^{2}-4(2)(15)}} {2(2)}

x=\frac{-11(+/-)\sqrt{121-120}} {4}

x=\frac{-11(+/-)\sqrt{1}} {4}

x=\frac{-11(+/-)1} {4}

x_1=\frac{-11(+)1}{4}=-2.5

x_2=\frac{-11(-)1}{4}=-3

therefore

The roots of the equation are x=-3 and x=-2.5

5 0
3 years ago
Carlton bought stock at $5.25 for each share. The price of the stock decreased $3.75 after one month. It then increased $1.85 an
Natali [406]

Answer:

The price of the stock is now $5.60

Step-by-step explanation:

$5.25 - $3.75 = $1.50

$1.50 + $1.85 + $2.25 = $5.60

8 0
3 years ago
Read 2 more answers
The area of a trapezoid is 60 square inches. One of the bases is 8 inches long. The height is 6 inches long. What is the length
Mashcka [7]

Answer:

12 in

Step-by-step explanation:

The area (A) of a trapezoid is calculated as

A = \frac{1}{2} h(b₁ + b₂ )

where h is the height and b₁, b₂ the parallel bases

Given h = 6, b₁ = 8 and A = 60 , then

\frac{1}{2} × 6 × (8 + b₂ ) = 60 , that is

3(8 + b₂ ) = 60 ( divide both sides by 3 )

8 + b₂ = 20 ( subtract 8 from both sides )

b₂ = 12

The length of the second base is 12 inches

6 0
3 years ago
Find the diameter of a circle with a circumference of 15.17 yards
Aleksandr-060686 [28]

Answer:

4.83

Step-by-step explanation:

To find the circumference of a circle you need to take the diameter times Pi. So if you need to find the diameter from the circumference, simply divide the circumference by 3.14 which is Pi

Hope this helps!

7 0
3 years ago
Trevor bought a sandwich for $4.05, a bag of chips for $2.05, and a drink for $1.60. The tax was $0.60. He gave the cashier $10.
jek_recluse [69]
He should receive $1.70 in change.

5 0
3 years ago
Other questions:
  • What is the answer to -3(1+6r)=14-r
    14·2 answers
  • Which lines are perpendicular to 5x + 4y = 8?
    7·2 answers
  • Combine the Expression:(10x - 4) + (-7 + x)
    13·2 answers
  • Barb is making a bead necklace she strings won't Whitebead then three blue beads that one white bead and Soren write the numbers
    15·1 answer
  • Please help me idk if I did it right
    12·1 answer
  • Litas softball team won 8 games last month and 10 this month, what was the percent change in game the team won? Was it an increa
    14·1 answer
  • Convert Decimal 98 to Binary<br> Pls help
    5·1 answer
  • Boat A leaves a dock headed due east at 2:00PM traveling at a speed of 9 mi/hr. At the same time, Boat B leaves the same dock tr
    13·1 answer
  • Henry ran 1 5/8 miles in the morning and 9/10 miles in the afternoon how many miles did he run in all
    12·2 answers
  • Help Me with this please
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!