All categories

3 years ago

2x+8y = 6 15x + 20y = 15

Mathematics

1 answer:

Likurg_2 [28]3 years ago

4 0

Answer:

The solution is (0, 3/4)

Step-by-step explanation:

Please copy and share the instructions. Here they are: Solve the following system of linear equations.

Both of the equations can be reduced (simplified):

2x+8y = 6 => x + 4y = 3

15x + 20y = 15 => 3x + 4y = 3

Let's use the elimination by addition and subtraction method. Multiply the first equation by -1, obtaining

-x - 4y = -3

Add the second 3x + 4y = 3

equation to the

first.

We get: 2x = 0.

Thus, x = 0. Substituting 0 for x in the 1st original equation yields:

2(0) + 8y = 6. Then y = 6/8, or y = 3/4.

The solution is (0, 3/4).

You might be interested in

Please hurry I will mark you brainliest <br><br> 8(3 - i) = 5(2 - 2i)

Mrac [35]

Answer:-7

Step-by-step explanation:

24-8i=10-10i

24-10=-10i+8i

14=-2i

I=-7

6 0

3 years ago

please help i am failing math class she said if you dont turn in this Backpack artifact task am going to fail it is 6th grade ma

agasfer [191]

Answer:

I might be able to help as long as there's no algebra involved

6 0

2 years ago

Read 2 more answers

Combine like terms. What is a simpler form of each expression?<br> 30 - 5d - C + 3d

borishaifa [10]

30 - 5d - c + 3d

In this equation there are two numbers which have the letter d with them. These would be like terms.

You have -5d + 3d which equals -2d

The simplified form would be 30 -2d -c

4 0

3 years ago

Read 2 more answers

In a bag of m&m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&m

olasank [31]

There are 5+6+4+3+2=20 m&m's in the bag.
Calculate in how many ways you can choose 3 m&m's from 20:
$_{20} C _3=\frac{20!}{3!(20-3)!}=\frac{20!}{3! \times 17!}=\frac{17! \times 18 \times 19 \times 20}{6 \times 17!}=\frac{18 \times 19 \times 20}{6}=3 \times 19 \times 20= \\
=1140$

There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
$_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20$

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
$P=
\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}$

The probability is 1/57.

4 0

3 years ago

Find the area and the circumference of a circle with diameter 9 yards using the value 3.14 for pi

Ray Of Light [21]

Area of circle = pi r^, (r)radius is half the diameter so 9÷2=4.5 so in this scenario it would be 3.14×4.5^
3.14×4.5×4.5=63.6

circumference of circle= diameter×pi (3.14)
so 9×3.14=28.3

5 0

3 years ago