MATHEMATICS HIGH SCHOOL

Write the equation of line perpendicular to y = 4 that passes through (5, 2).

Answers

Answer 1

Answer:

Answer:

x= -5

Step-by-step explanation:

just graphed it . y = 4 is a flat horizontal line.

x=5 is a vertical line that passes through (5,2)

Answer 2

Answer:

Final answer:

The equation of the perpendicular line is x = 5.

Explanation:

To find the equation of a line perpendicular to y = 4 that passes through the point (5, 2), we need to determine the slope of the original line. The equation y = 4 represents a horizontal line with a slope of 0. Perpendicular lines have slopes that are negative reciprocals of each other. The negative reciprocal of 0 is undefined, so the equation of the perpendicular line will be x = 5.

Learn more about Perpendicular lines here:

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ANSWER

1. 54

2. 4

3. 2

4. 60

5.46

Answer:

1. 54

2. 4

3. 2

4. 60

5.46

Activity 4: Factor Then solve1. x^2+7x=0
2. 6s^2+18s=0
3. t^+8t+16=0
4. x^2-10x+25=0
5. h^2+6h=16
6. x^2-14=5x
7. 11r+15=-2r^2
8. x^2-25=0
9. 81-4x^2=0
10. 4s^2+9=12s

PLEAAASE HELP ME!

Answers

very complex, you should ask each individual question, or give more points
general strategy:
get 1 side to equal 0
factor
set each factor to 0
solve

1. factor out the x and set each to zero
x(x+7)=0
x=0

x+7=0
subtract 7
x=-7
x=0 or -7

2. factor out the 6s
6s(s+3)=0
6s=0
s=0

s+3=0
subtract 3
s=-3
s=-3 or 0

3. I'm assuming that the equation was supposed to be t^2+8t+16 so
find what 2 numbers multiply to get 16 and add to get 8
the numbers are 4 and 4
factor
(x+4)(x+4)=0
x+4=0
subtract 4
x=-4

4. find what 2 numbers multipy to get 25 and add to get -10
the numbers are -5 and -5
factor
(x-5)(x-5)=0
set each to zero
x-5=0
add 5
x=5

5. subtract 16 from both sides
h^2+6h-16=0
find what 2 numbers multiply to get -16 and add to get 6
the numbers are -6 and 10
factor
(h-6)(h+10)=0
set each to zero
h-6=0
add 6
h=6

h+10=0
subtract 10
h=-10
h=-10 or 6

6.subtract 5x from both sides
x^2-5x-14=0
find what 2 numbers multiply to get -14 and add to get -5
the numbers are -7 and 2
factor
(x-7)(x+2)=0
set each to zero
x-7=0
add 7
x=7

x+2=0
subtract 2
x=-2
x=-2 or 7

7.add 2r^2 to both sides
2r^2+11r+15=0
difficult so multiply 2 and 15 and get 30
not factor and find what 2 numbers multiply to get 30 and add to get 11
the numbers are 5 nd 6
split 11 up into 5 and 6
2r^2+6r+5r+15=0
group and factor
(2r^2+6r)+(5r+15)=0
factor
2r(r+3)+5(r+3)=0
reverse distribute
(2r+5)(r+3)=0
set each to zero
2r+5=0
subtract 5
2r=-5
divide 2
r=-5/2

r+3=0
subtract 3
r=-3
r=-3 or -5/2

8. this is a difference of 2 perfect squares in which if you have
x^2-y^2 that factors out of (x-y)(x+y) so
x^2-25=0
x^2-5^2=0
(x-5)(x+5)=0
set each to zero
x-5=0
add 5
x=5

x+5=0
subtract 5
x=-5
x=-5 or 5

9. differnce of 2 perfect squares
81-4x^2=0
9^2-(2x)^2=0
(9-2x)(9+2x)=0
set each to zero
9-2x=0
add 2x to both sides
9=2x
divde 2
9/2=x

9+2x=0
subtract 9 from both sidees
2x=-9
divide 2
x=-9/2
x=-9/2 or 9/2

10. subtract 12s from both sides
4s^2-12s+9=0
multiply 4 and 9 and get 36
find what 2 numbers multiply to get 36 and add to get -12
the numbers are -6 and -6
split up the -12s into -6s and -6s
4s^2-6s-6s+9=0
group
(4s^2-6s)+(-6s+9)=0
factor
2s(2s-3)+(-3)(2s-3)=0
reverse distribute
(2s-3)(2s-3)=0
set each to zero
2s-3=0
add 3
2s=3
divide 2
s=3/2

ANWERS:
1. x=0 or -7
2. s=-3 or 0
3. x=-4
4. x=5
5. h=-10 or 6
6. x=-2 or 7
7. r=-3 or -5/2
8. x=-5 or 5
9. x=-9/2 or 9/2
10. s=3/2

There are 15 students in a social studies class. Two students will be selected to present their term projects today. In how many different orders can two students be selected?

Answers

Given:
15 students
2 students must be chosen.
No repetition, no order

This is a combinations problem. We use this formula: n! / (n-r)!(r!)

n = 15 ; r = 2

15! / (15-2)!(r!) ⇒ 15! / 13! * 2! = 105

Using the combination formula, it is found that the students can be selected in 105 ways.

The order in which the students are selected is not important, hence the combination formula is used to solve this question.

What is the combination formula?

The number of possible combinations of x elements from a set of n elements is given by:

$C_((n,x)) = (n!)/(x!(n-x)!)$

In this problem, two students are selected from a set of 15, hence the number of different orders is given by:

$C_(15,2) = (15!)/(2!13!) = 15 * 7 = 105$

More can be learned about the combination formula at brainly.com/question/25821700

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On a shopping trip, Peter spent 1/3 of his money for a jacket and another $5 for a hat. If peter still had 1/2 of his money left, how much money did he have originally ??a) $18, b) $24, c) $30, d) $48, e) $60

Answers

(a) 18
18 / 3 = 6
6 + 5 = 11
18 - 11 = 7
18 / 2 = 9
$7 \neq 9$

(b) 24
24 / 3 = 8
8 + 5 = 13
24 - 13 = 11
24 / 2 = 12
$11 \neq 12$

(c) 30
30 / 3 = 10
10 + 5 = 15
30 - 15 = 15
30 / 2 = 15
$15 = 15$

(d) 48
48 / 3 = 16
16 + 5 = 21
48 - 21 = 27
48 / 2 = 24
$27\neq 24$

(e) 60
60 / 3 = 20
20 + 5 = 25
60 - 25 = 35
60 / 2 = 30
$35 \neq 30$

Answer: (c) $30

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