Could someone check my Algebra work?

Question

Could someone correct my answers and help with the other problems?

1. Are the following lines parallel, perpendicular, or neither? L1 with equation x – 5y = 10, L2 with equation 5x + y = 5.

Answer: Neither (I'm not to sure but i think this is the right answer)

2. Divide: a^10/a^4

How would I complete this problem?

3. A rectangle has sides of 3x – 4 and 7x + 10.Find the expression that represents its perimeter.

Answers: Answer: P = 20x + 12

4. Write the equation of the line that passes through point (–6, 7) with a slope of 0.

Answer: y = 7

5. Find the slope of the line passing through the points (1, 1) and (3, 3).

Answer: 1

6. Evaluate. 5 • 4 ÷ 2 – 42

Answer: -6

7. Find the slope and the y-intercept.

y = x

Answer: Slope is 1, y-intercept (I'm not sure what that answer is)

Answer 1

1. Are the following lines parallel, perpendicular, or neither? L1 with equation x – 5y = 10, L2 with equation 5x + y = 5.

Answer: change equation to standard form y = mx + b
L1 y = x/5 - 2 (slope = 1/5)
L2 y = 5 - 5x (slope = -5)
Perpendicular

2. Divide: a^10/a^4
Subtract the exponents
a^10 / a^4 = a^(10-4) = a^6

3. A rectangle has sides of 3x – 4 and 7x + 10.Find the expression that represents its perimeter.

Answers: P = 20x + 12
Correct!

4. Write the equation of the line that passes through point (–6, 7) with a slope of 0.

Answer: y = 7
Correct

5. Find the slope of the line passing through the points (1, 1) and (3, 3).

Answer: 1
Correct

6. Evaluate. 5 • 4 ÷ 2 – 42

Answer: -6
5*4 / 2 - 42 = 20/2 - 42 = 10 - 42 = -32

7. Find the slope and the y-intercept.
y = x

Answer: Slope is 1, y-intercept
y = mx + b (m is slope and b = y-intercept)
y = x + 0
slope = 1; y-intercept = 0

Answer 2

1. Are the following lines parallel, perpendicular, or neither? L1 with equation x – 5y = 10, L2 with equation 5x + y = 5.

Answer: Neither (I'm not to sure but i think this is the right answer)

Start by putting them in slope-intercept form:
L1 y = (1/5)x - 2
L2 y = -5x + 5

The are perpendicular. Perpendicular lines have slopes that are negative reciprocals of each other

2. Divide: a^10/a^4

How would I complete this problem?

There are a few ways I could show this but let's use laws of exponents:
a^10/a^4 = a^10*a^(-4) = a^(10-4) = a^6

3. A rectangle has sides of 3x – 4 and 7x + 10.Find the expression that represents its perimeter.

Answers: Answer: P = 20x + 12

Since it is a rectangle it has 2 pairs of sides of equal lengths:
2(3x - 4) + 2 (7x + 10)
6x - 8 + 14x + 20
20x + 12
So you were correct. :)

4. Write the equation of the line that passes through point (–6, 7) with a slope of 0.

Answer: y = 7

right: y = 0x + 7
y = 7

5. Find the slope of the line passing through the points (1, 1) and (3, 3).

Answer: 1

slope may be found by rise/run
(y2-y1)/(x2-x1)
(3-2)/(3-2)
1/1 = 1 yes. :)

6. Evaluate. 5 • 4 ÷ 2 – 42

Answer: -6
PEMDAS:
20/2 - 42
10 - 42
-32

Remember your order of operations

7. Find the slope and the y-intercept.

y = x

Answer: Slope is 1, y-intercept (I'm not sure what that answer is)
Let's look at the slope-intercept general form:
y = mx + b compared to:
y = x

The coefficient of x is 1 and b = 0
Thus the slope is 1. The y intercept is 0.

Answer 3

1. Perpendicular. In slope-intercept form L1's equation is y = 1/5x - 2, and L2's is y = -5x + 5.

1/5 and -5 are negative reciprocals, so lines with those slopes are perpendicular.

2. a^6. Subtract exponents when dividing.

3. 20x+12 is correct

4. y=7 is correct

5. 1 is correct

6. I get -32 (5*4 = 20, 20/2 = 10, 10-42=-32)

7. Slope =1. Y-intercept is the value of y when x=0. Since y=x, y=0 when x=0, and the y-intercept is 0.

Answer 4

1- the lines are perpendicular - if you solve for y in both equations you would see that the slopes would be x/5 and -5. these two slopes are opposite reciprocals and when the lines have the slopes that are opposite reciprocals, then they are perpendicular.

2-a^10/ a^4 is equal to a^6 - to do this, all you have to do is subtract the powers. Since the bases are the same, the way to solve the equation is the subract the powers.

3- sorry, I don't know the answer for this question.

4- your answer is right

5- your answer is right

6- for this one, I got -32, not -6. - to do this problem, you have to follow the order of operations. first to multiply 5 * 4= 20, then you divide it by 2 = 10, then you subract it from -42. The answer would be -32.

7-the slope is 1 and the y-intercept is at the point of (0.0)

Answer 5

1st they are perpendicular as m1*m2= -1
2nd a^(10-4)=a^6
3rd correct
4th correct
5th correct
6th incoorect
According to bodmas rule the problem becomes 5*(4/2)-42
=5*2-42
=(5*2)-42
=10-42
=-32
Note the brackets have to be solved first
7th slope is 1 and y intercept is 0 as c=0 in y=mx+c

Answer 6

1.
x – 5y = 10
-5y = -x + 10
y = 0.5x - 2

5x + y = 5.
y = -5x +5

Same slope, lines are perpendicular

2.
a^10/a^4
= a^(10 - 4)
= a^6

3.
P = 2(3x – 4) + 2(7x + 10)
P = 6x - 8 + 14x + 20
P = 20x + 12

4.
Equation: y-y0 = m(x-x0)
(x0,y0) = (-6,7)
m = 0
y - 7= 0
y = 7
you're right

5.
m = (y2 - y1)/(x2-x1)
(x1,y1) = (1,1)
(x2,y2) = (3,3)

m = 2/2 = 1
you're right

6.
5 • 4 ÷ 2 – 42
= (5*4/2) - 42
= (10) - 42
= -32

7.
Equation: y = mx + b
m: slope
b: y-intercept

y = x
m = 1
b = 0
(y-intercept is zero)

Answer 7

1.)perpendicular (just worry about the coefficient in front of the x after u've put it into slope intercept form (y=mx+b))
2.) a^6
3.)ur right
4.)ur right
5.)ur right
6.)[(20)/2]-42=-32 (order of operations)
7.)y=mx+b The slope is 1 and the y intercept is zero.

and to nikhil, you can go **** bodma