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lorasvet [3.4K]
4 years ago
9

Number of cars Repaired at

Mathematics
1 answer:
sveta [45]4 years ago
6 0
What if this !? Tricky answer : explanation
You might be interested in
Factor 1 - 2.25x^8.
sleet_krkn [62]
<span>1 - 2.25x</span>⁸ =
1² - (1.5x⁴)² =
(1 + 1.5x⁴)(1 - 1.5x⁴)
5 0
3 years ago
Read 2 more answers
Just need help on #7 thru 8(a), b and c
Stella [2.4K]

7

a. The slope of the line passing through the points R(3, 5) and H(-1,2) is -3/4

b. The distance between two points (x₁, y₁) and (x₂,y₂) is 5 units

c.  The midpoint of the R(3, 5) and H(-1,2) is  (2, 4)

8.

a. The transformation rule is  (x,y) → (x + 1, y - 1)

b.

  • The x-coordinate is shifted 1 unit to the left and
  • The y - coordinate is shifted 1 unit downwards.

c. The image of B' the pre-image of B(5, 6) is (6, 5)

<h3 /><h3>7 a. How to find the slope of the line?</h3>

The slope of a line passing through the points (x₁, y₁) and (x₂,y₂) is m = (y₂ - y₁)/(x₂ - x₁)

Given that

  • (x₁, y₁) = (3, 5) and
  • (x₂,y₂) = (-1, 2)

So, m = (y₂ - y₁)/(x₂ - x₁)

m = (2 - 5)/(-1 - 3)

m = -3/-4

m = -3/4

So, the slope of the line passing through the points R(3, 5) and H(-1,2) is -3/4

<h3>b. The distance between the points</h3>

The distance between two points (x₁, y₁) and (x₂,y₂) is d = √[(y₂ - y₁)² + (x₂ - x₁)²]

Given that

  • (x₁, y₁) = (3, 5) and
  • (x₂,y₂) = (-1, 2)

d = √[(y₂ - y₁)² + (x₂ - x₁)²]

d = √[(2 - 5)² + (-1 - 3)²]

d = √[(-3)² + (-4)²]

d = √[9 + 16]

d = √25

d = 5 units

So, the distance between two points (x₁, y₁) and (x₂,y₂) is 5 units

<h3>7 c How to find the midpoint of the R(3, 5) and H(-1,2) </h3>

The midpoint of the the points (x₁, y₁) and (x₂,y₂) is (x, y)  = [(x₁ + x₂)/2, (y₁ + y₂)/2]

Given that

  • (x₁, y₁) = (3, 5) and
  • (x₂,y₂) = (-1, 2)

So, the midpoint (x, y)  = [(x₁ + x₂)/2, (y₁ + y₂)/2]

(x, y)  = [(3 + (-1))/2, (5 + 3)/2]

(x, y)  = [(3 - 1)/2, (5 + 3)/2]

(x, y)  = [4/2, 8/2]

(x, y)  = (2, 4)

So, the midpoint of the R(3, 5) and H(-1,2) is  (2, 4)

<h3>8. a The rule for the transformation of point A(1, 4) to point B(2, 3)</h3>

Given that point A(1, 4) and point B(2, 3) we see that point B(1 + 1, 4 - 1).

Let pont A be (x,y).

So, point B = (x + 1, y - 1)

So, the transformation rule is  (x,y) → (x + 1, y - 1)

<h3>b. Describe the transformation</h3>

Since the transformation rule is   (x,y) → (x + 1, y - 1), we see that

  • the x-coordinate is shifted 1 unit to the left and
  • the y - coordinate is shifted 1 unit downwards.
<h3>c. The image of B' the pre-image of B(5, 6)</h3>

Since the transformation rule is  (x,y) → (x + 1, y - 1) and point B is (5, 6), thus the image of B' is

(x,y) → (x + 1, y - 1)

(5,6) → (5 + 1, 6 - 1)

(5,6) → (6, 5)

So, the image of B' the pre-image of B(5, 6) is (6, 5)

Learn more about slope of a line here:

brainly.com/question/1617757

#SPJ1

3 0
2 years ago
I need a normal answer not a equation
Sloan [31]

Answer:

PQ = 3.2

Step-by-step explanation:

You must use the addition property of equality, and subtract 16 from the of the 5x, and add it to the other side.

After that, you have 5x = 16

You then divide both sides by 5.

5x/5 = 1x = x

16/5 = 3.2

x = 3.2

Therefore, the side 'PQ' is equal to 3.2

Brainliest appreciated!

4 0
3 years ago
For which of the following equations are x = 5 and x = –5 both solutions?
Minchanka [31]
Short Answer B
Argument
A
A will give you x = +/- 5i 
x^2 + 25 = 0
x^2 = - 25       Take the square root.
sqrt(x^2) = +/- sqrt(-25)
x = +/- (5)i which is a complex number.


B
Is the answer
x^2 = 25
sqrt(x)^2 = sqrt(25)
x = +/- 5

C
Can't be factored just by looking at it. You can show that C is not true just by putting 5 into the equation
f(x) = x^2 + 10x - 25
f(5) = 25 + 10*5 - 25
f(5) = 50
C is not true.

D
D can be eliminated as C was
f(x) = x^2 - 5x - 25
f(5) = -25 ( l'll let you show this is not true). 5 is not a solution because it does not make f(x) = 0
6 0
3 years ago
Read 2 more answers
Hi ! How to solve this question?
tester [92]

You should actually have

-x^2-4x-10=-(x^2+4x+10)=-(x^2+4x+4+6)=-((x+2)^2+6)=-(x+2)^2-6

Now, remember that x^2 is always non-negative, so (x+2)^2\ge0 for any value of x. This means -(x+2)^2\le0 for any x, and so

-(x+2)^2-6\le0-6=-6

i.e. f(x) is at most -6, and hence negative for all x.

4 0
3 years ago
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