A. Alright, we want to multiply one equation by a constant to make it cancel out with the second. Since the first equation has a "blank" y, let's multiply the first equation by <em>2</em>.

3x-y=0 → 2(3x-y=0) = 6x - 2y = 0

5x+2y=22

The answer for this part would be: 6x - 2y = 0 and 5x + 2y = 22

B. So now we combine them:

6x - 2y = 0

+ + +

5x + 2y = 22

= = =

11x + 0 = 22 ← The answer

C. Now that we have the equation 11x = 22, we solve for x

11x = 22 ← Divide both sides by 11

x = 2 ← The answer

D. Now that we have x=2, we plug that back in to 5x+2y=22 and solve for y:

5(2)+2y = 22

10 + 2y = 22

2y = 12

y = 6

<u>Therefore, the solution to this problem is x = 2 and y = 6</u>

5 0

3 years ago

If minor arc AB measures 9pi, what is the length of the radius of circle C? If necessary, round your answer to the nearest inch.

Y_Kistochka [10]

Answer: 12 inches

Step-by-step explanation:

took the test

8 0

3 years ago

Read 2 more answers

HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

3 years ago

The least common multiple of two numbers is 60 and one of the numbers is 7 less than the other number.What are the numbers.Justi

denis-greek [22]

Answer:

As the least common multiple of two numbers is 60 , the two numbers are factors of 60 . Among {1,2,3,4,5,6,10,12,15,20,30,60} , 3 & 10 and 5 & 12 are the only two pair of numbers whose difference is 7 . But Least common multiple of 3 and 10 is 30

6 0

3 years ago