Match the system of linear equations with its graph. Then determine whether the system has one solution, no solution, or infinitely many solutions. 3.-x+y=1 x-y=1 5.2x+y=4 -4x-2y=-8 7.-2x+4y=1 3x-6y=9

Answers

Answer 1

Answer:

The system of linear equations that match with its graph and the solution is

Number 3 with F has no solution.
Number 4 with E has infinitely many solutions.
Number 5 with B infinitely many solutions.
Number 6 with C has one solution.
Number 7 with D has no solution.
Number 8 with A has one solution.

Making a graph of a system of linear equations is done by substituting the value x = 0 and the value y = 0 in an equation. After getting two points, draw a line through the two coordinates. But if the equation goes past the point (0 , 0) then take another value of x at random to get another point. Graph and solution

The parallel lines will have no solution.
Two intersecting lines will have one solution.
Two coincide lines will have infinitely many solutions.

3. - x + y = 1

x = 0
0 + y = 1
y = 1
(0 , 1)
y = 0
- x + 0 = 1
x = - 1
(- 1 , 0)

x - y = 1

x = 0
0 - y = 1
y = - 1
(0 , - 1)
y = 0
x - 0 = 1
x = 1
(1 , 0)

Graph F matches the system.

The two lines are parallel.

No solution.

4. 2x - 2y = 4

x = 0
- 2y = 4
y = - 2
(0 , - 2)
y = 0
2x = 4
x = 2
(2 , 0)

- x + y = - 2

x = 0
y = - 2
(0 , - 2)
y = 0
- x = - 2
x = 2
(2 , 0)

Graph E matches the system.

The two lines coincide.

Infinitely many solutions.

5. 2x + y = 4

x = 0
y = 4
(0 , 4)
y = 0
2x = 4
x = 2
(2 , 0)

- 4x - 2y = - 8

x = 0
- 2y = - 8
y = 4
(0 , 4)
y = 0
- 4x = - 8
x = 2
(2 , 0)

Graph B matches the system.

The two lines coincide.

Infinitely many solutions.

6. x - y = 0

x = 0
- y = 0
y = 0
(0 , 0)
x = 2
2 - y = 0
y = 2
(2 , 2)

5x - 2y = 6

x = 0
- 2y = 6
y = - 3
(0 , - 3)
y = 0
5x = 6
x = 6/5
(6/5 , 0)

Graph C matches the system.

The two lines intersect.

One solution.

7. - 2x + 4y = 1

x = 0
4y = 1
y = 1/4
(0 , 1/4)
y = 0
- 2x = 1
x = - 1/2
(- 1/2 , 0)

3x - 6y = 9

x = 0
- 6y = 9
y = - 3/2
(0 , - 3/2)
y = 0
3x = 9
x = 3
(1 , 3)

Graph D matches the system.

The two lines are parallel.

No solution.

8. 5x + 3y = 17

x = 0
3y = 17
y = 17/3 = 5 2/3
(0 , 5 2/3)
y = 0
5x = 17
x = 17/5 = 3 2/5
(3 2/5 , 0)

x - 3y = - 2

x = 0
- 3y = - 2
y = 2/3
(0 , 2/3)
y = 0
x = - 2
(- 2 , 0)

Graph A matches the system.

The two lines intersect.

One solution.

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A sample of radioactive waste has a half-life of 10 years and an activity level of 2 curies. After how many years will the activity level of this sample be 0.25 curie?a. 10 yearsb. 20 yearsc. 30 yearsd. 40 yearse. 80 years

Answers

We can use formula for half life of radioactive substance to get the needed time interval's length.

Option: C: 30 years.

After 30 years, the activity level of this sample will be 0.25

Given that:

Length of half life = b = 10 years
Activity level = 2 curies

Let after y years, the activity level of this sample be 0.25.

What is the relationship between half life and amount remaining?

$(\lambda)/(\lambda _1) = 2^{(a)/(b)}$ ,

where λ₁ = remainingamount of sample, λ = original amount of sample, a = total time length for disintegration, b = halflife.

How to find the remaining time for sample to be with 0.25 activity level?

Putting values

$\lambda = 2 \: \rm curie\n\lambda _1 = 0.25 \: \rm curie\nb = 10$

$(2)/(0.25) = 2^{(a)/(10)}\n\n(a)/(10) = 3\n\na = 30$

Thus, after 30 years, the activity level of this sample will be 0.25

Thus, option C : 30 years is correct.

Learn more about half life of radioactive substances here:

brainly.com/question/13234158

Answer:

c. 30 years

Step-by-step explanation:

The expression for half life of a radioactive substance is given as

λ/λ₁ = 2⁽ᵃ/ᵇ⁾.............. Equation 1

Where λ₁ = remaining sample, λ = original sample, a = Total disintegration time, b = half life.

Given: λ = 2 curies, λ₁ = 0.25 curies, b = 10 years.

Substituting these values into equation 1

2/0.25 = 2⁽ᵃ/¹⁰⁾

8 = 2⁽ᵃ/¹⁰⁾

2³ = 2⁽ᵃ/¹⁰⁾

Since the base in both side of the equation is the same, we can equate the base.

3 = a/10

a/10 = 3

a = 10×3

a = 30 years.

Thus it will take 30 years for the activity level of the sample to be 0.25 curie

The right option is c. 30 years

Please Help me solve: 7/8 * 1/4
Please give your answer in simplest form, Thank You!

Answers

Make them the same demonatator so 7/8 times 2/8 which equals 14 over 8 for a mixed number of 1 6/8

A hot air balloon is flying above Groveburg. To the left side of the balloon, the balloonist measure the angle of depression to the Groveburg soccer fields to be 20° 15'. To the right side of the balloon, the balloonist measures the angle of depression to the high school football field to be 62° 30'. Find the distance from the balloon to the soccer fields

Answers

Answer:

Solution : 3.6 miles

Step-by-step explanation:

The first step in solving this problem is to convert 15' to degrees --- (1) . 15' is represented as 15 minutes, so to convert to degrees we have 15 / 60 = 0.25°. Adding 20 + 15' we should get 20 + 0.25 = 20.25 (degrees).

Now remember that on the right side we have the football field, 62° 30'. Let's convert that 30' into degrees --- (2). 30 / 60 = 0.5, and 62 + 0.5 = 62.5.

Using both this information we can calculate the angle from the balloon to the horizontal. That would be 180 - 20.25 - 62.5 = 97.25°. Therefore the distance from the balloon to the soccer fields would be as follows,

distance / sin(62.5) = 4 / sin(97.25)

Let distance = x ...

x / sin(62.5) = 4 / sin(97.25),

x / 0.88701083317 = 4 / 0.99200494968,

x * 0.99200494968 = 4 * 0.88701083317,

x * 0.99200494968 = 3.548,

x = ( About ) 3.5766 miles

Given the options the distance from the balloon to the soccer fields would be 3.6 miles.

Simplify the expression: 1/1+cot^2xa.sec^2x
b.csc^2x
c.sin^2x
d.cos^2x
e.tan^2x

Answers

$\sin ^( 2 ){ x } +\cos ^( 2 ){ x } =1\n \n \frac { \sin ^( 2 ){ x } }{ \sin ^( 2 ){ x } } +\frac { \cos ^( 2 ){ x } }{ \sin ^( 2 ){ x } } =\frac { 1 }{ \sin ^( 2 ){ x } } \n \n 1+\cot ^( 2 ){ x } =\csc ^( 2 ){ x }$

Because of this...

$\frac { 1 }{ 1+\cot ^( 2 ){ x } } \n \n =\frac { 1 }{ \csc ^( 2 ){ x } }$

But...

$\frac { 1 }{ \csc ^( 2 ){ x } } =\sin ^( 2 ){ x }$

Therefore:

$\frac { 1 }{ 1+\cot ^( 2 ){ x } } =\sin ^( 2 ){ x }$

Answer:

(c)

(x + 5)2 + (y - 7)2 = 36, its radius is

Answers

The answer would be 6

Steps: The equation of a circle is (x – h)² + (y – k)² = r²; since we are trying to find the radius we are going to focus on the r². Now if you look the equation has the 36 filled in as the r² so r² = 36; we now have to isolate the r to get the radius, how do we get the radius? We Square root the r² and 36. After we get r = 6, The radius is 6 units

Please help I need this for a math midterm the question is below in the picture !!

Answers

Answer:the first answer is store a. the second one median is shorter.then store b then store c

Step-by-step explanation:

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