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3241004551 [841]
3 years ago
9

Do all graphs of linear equations in two variables have a y intercept?

Mathematics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer:

Yes and no because not all are in the y intercept it depends

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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5 0
3 years ago
Solve the following system of equations.
Yuri [45]

Answer:

x= 12

y= 10

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7 0
3 years ago
A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)
DochEvi [55]

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

Speed=\frac{distance}{time}

in this problem we have

distance=400\ m \\time=120\ sec

substitute in the formula

Speed=\frac{400}{120}

Speed=3.5\frac{m}{sec}

therefore

<u>the answer Part a) is</u>

the speed is equal to 3.5\frac{m}{sec}

<u>Part b) </u>Find the velocity

we know that

<u>Velocity </u>is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

magnitude=3.5\frac{m}{sec}

direction=North\ East\ (NE)

therefore

<u>the answer Part b) is</u>

the velocity is 3.5\frac{m}{sec}\ North\ East\ (NE)

Part c)

we know that

the acceleration is equal to the formula

a=\frac{V2-V1}{t2-t1}

in this problem we have

V2=0 \\V1=3.5\frac{m}{sec}

t2=15\ sec\\t1=0

substitute in the formula

a=\frac{0-3.5}{15-0}

a=-\frac{3.5}{15}\frac{m}{sec^{2}}

a=-\frac{7}{30}\frac{m}{sec^{2}}

therefore

<u>the answer Part c) is</u>

the acceleration is -\frac{7}{30}\frac{m}{sec^{2}}

This is an example of negative acceleration

7 0
3 years ago
Read 2 more answers
Complete the table of values for this absolute value function. Then use the drawing tool(s) to graph the function. F(x)-2|x+1|-1
rusak2 [61]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the equation

f\left(x\right)=-2|x+1|-1

As some of the absolute rules are:

  • \left|a\right|=a,\:a\ge 0
  • \:|-a|=a,\:\quad \:a>0

NOW, let us solve!

Let us substitute all the table values

Putting x = -4

y=-2\left|-4+1\right|-1        ∵\mathrm{Apply\:absolute\:rule}:\quad \left|a\right|=a,\:a\ge 0

y=-6-1

y=-7

So, when x = -4, then y = -7

Putting x = -3

y=-2|-3+1|-1\:

y=-4-1

y=-5

when x = -3, then y = -5

Putting x = -2

y=-2\left|-2+1\right|-1

y=-2-1

y=-3

when x = -2, then y = -3

Putting x = -1

y=-2\left|-1+1\right|-1

y=-0-1

y=-1

when x = -1, then y = -1

Putting x = 0

y=-2\left|0+1\right|-1

y=-2-1

y=-3

when x = 0, then y = -3

Putting x = 1

y=-2\left|1+1\right|-1

y=-4-1

y=-5

when x = 1, then y = -5

The graph is also attached below.

6 0
3 years ago
Which arc is a semicircle?
hichkok12 [17]

Answer:

BE is the semicircle

7 0
3 years ago
Read 2 more answers
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