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lys-0071 [83]
3 years ago
8

Which of these lines have a scale factor of 2?

Mathematics
2 answers:
Natalka [10]3 years ago
5 0
The answer would most likely be A
guajiro [1.7K]3 years ago
4 0

Answer:

The answer will be A

Step-by-step explanation:

You might be interested in
Which Funtion is represented by the graph below? PLS PLS HELP DO TMR
igor_vitrenko [27]

Answer:

straight line ,

Step-by-step explanation:

because you can see that the line is straight ,

4 0
3 years ago
I desperately need this question answered. See attached image thanks so much
Dahasolnce [82]

For this case we must follow the steps below:

step 1:

We place each of the given points on a coordinate axis

Step 2:

We join the AC points (represented by the orange line)

We join the BD points (represented by the blue line)

It is observed that the resulting figure after placing the 4 points on a coordinate axis, turns out to be a rhombus.

In addition, the blue and orange lines turn out to be perpendicular, that is, they have an angle of 90 degrees between them. This can be verified by finding the slopes of each of the two straight lines (blue and orange), which must be opposite reciprocal, that is, they comply: m1 * m2 = -1

In this case, the slope of the orange line is m1 = 1 and that of the blue line is m2 = -1

Then m1 * m2 = 1 * -1 = -1, it is verified that they are perpendicular.

Thus, the conclusion is that ABCD is a rhombus and AC is perpendicular to BD.

Answer:

See attached image

Option D

6 0
3 years ago
WILL MARK BRAINLIEST. Can someone help me wit des 2 questions?
Mnenie [13.5K]

Answer:

Zero is a number that can be equal to its opposite.

So, the given equation has solution for which LHS=RHS=0.

Step-by-step explanation:

The answer to the equation is 2

7 0
3 years ago
1.What are the zeros of the polynomial function?
Lorico [155]
Let's to the first example:

f(x) = x^2 + 9x + 20

Ussing the formula of basckara

a = 1
b = 9
c = 20

Delta = b^2 - 4ac

Delta = 9^2 - 4.(1).(20)

Delta = 81 - 80

Delta = 1

x = [ -b +/- √(Delta) ]/2a

Replacing the data:

x = [ -9 +/- √1 ]/2

x' = (-9 -1)/2 <=> - 5

Or

x" = (-9+1)/2 <=> - 4
_______________

Already the second example:

f(x) = x^2 -4x -60

Ussing the formula of basckara again

a = 1
b = -4
c = -60

Delta = b^2 -4ac

Delta = (-4)^2 -4.(1).(-60)

Delta = 16 + 240

Delta = 256

Then, following:

x = [ -b +/- √(Delta)]/2a

Replacing the information

x = [ -(-4) +/- √256 ]/2

x = [ 4 +/- 16]/2

x' = (4-16)/2 <=> -6

Or

x" = (4+16)/2 <=> 10
______________

Now we are going to the 3 example

x^2 + 24 = 14x

Isolating 14x , but changing the sinal positive to negative

x^2 - 14x + 24 = 0

Now we can to apply the formula of basckara

a = 1
b = -14
c = 24

Delta = b^2 -4ac

Delta = (-14)^2 -4.(1).(24)

Delta = 196 - 96

Delta = 100

Then we stayed with:

x = [ -b +/- √Delta ]/2a

x = [ -(-14) +/- √100 ]/2

We wiil have two possibilities

x' = ( 14 -10)/2 <=> 2

Or

x" = (14 +10)/2 <=> 12
________________


To the last example will be the same thing.

f(x) = x^2 - x -72

a = 1
b = -1
c = -72

Delta = b^2 -4ac

Delta = (-1)^2 -4(1).(-72)

Delta = 1 + 288

Delta = 289

Then we are going to stay:

x = [ -b +/- √Delta]/2a

x = [ -(-1) +/- √289]/2

x = ( 1 +/- 17)/2

We will have two roots

That's :

x = (1 - 17)/2 <=> -8

Or

x = (1+17)/2 <=> 9


Well, this would be your answers.


7 0
3 years ago
The graph below describes the journey of a train between two cities.
Veseljchak [2.6K]

Answer:

[A] Acceleration in first 15 min = 200  km/h²

[B] Distance between two cities = 125 km

[C] Average Speed of journey =  62.5  km/hr

Step-by-step explanation:

To Solve:

Acceleration in first 15 min

Distance between two cities  

Average speed of journey

Solve:

[A] Work out the acceleration, in km/h?, in the first 15 mins.

Each horizontal block is 1/8 hr = 7.5 min

Each vertical block is 10 km/hr

Time                     Velocity  km/hr

0 Min  ( 0 hr)            0

15 Min (1/4 hr)           50

45 Min (3/4 hr)         50

60 MIn  ( 1 hr)            100

90 Min  ( 3/2 hr)        100

120 Min ( 2hr)            0

Acceleration in first 15 min  (1/4 hr)  =  (50 - 0)/(1/4 - 0)  = 50/(1/4)

= 200  km/h²

Hence, Answer for [A] = 200  km/h²

[B] Work out the distance between the two cities. $60 Velocity (in km/h)

= (1/2)(0 + 50)(1/4 - 0)  + 50 * (3/4 - 1/4)  + (1/2)(50 + 100)(1 - 3/4)  + 100 * (3/2 - 1) + (1/2)(100 + 0)(2 - 3/2)

=  25/4  + 25 + 75/4   + 50 + 25

= 125

Distance between two cities = 125 km

[C] Work out the average speed of the train during 20 the journey.

Average Speed of journey = 125/2  = 62.5  km/hr

[RevyBreeze]

5 0
2 years ago
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