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3241004551 [841]

3 years ago

9

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

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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Theo manufactures glass bottles that hold 12 fluid Ounces, with a tolerance of 0.2 fluid ounces. Write an inequality that can be

Lynna [10]

Answer:

23325435

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

best of luck mate

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

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