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AlladinOne [14]

2 years ago

13

Draw a mapping diagram (1,0) (3,0) (5,4) (7,4) (9,4)

1 answer:

Anton [14]2 years ago

3 0

Answer:

See attachment

Step-by-step explanation:

In other to draw a mapping diagram for the ordered pairs (1,0) (3,0) (5,4) (7,4) (9,4).

We identify the domain and range:

The domain is the set of the first coordinates of the ordered pairs.

Domain={1,3,5,7,9}

The range is the set of all the second coordinates.

Range: {0,4}

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Give the following equations determine if the lines are parallel perpendicular or neither

GaryK [48]

In order to determine whether the equations are parallel, perpendicular, or neither, let's simply each equation into a slope-intercept form or basically, solve for y.

Let's start with the first equation.

$\frac{6x-5y}{2}=x+1$

Cross multiply both sides of the equation.

$6x-5y=2(x+1)$ $6x-5y=2x+2$

Subtract 6x on both sides of the equation.

$6x-5y-6x=2x+2-6x$ $-5y=-4x+2$

Divide both sides of the equation by -5.

$-\frac{5y}{-5}=\frac{-4x}{-5}+\frac{2}{-5}$ $y=\frac{4}{5}x-\frac{2}{5}$

Therefore, the slope of the first equation is 4/5.

Let's now simplify the second equation.

$-4y-x=4x+5$

Add x on both sides of the equation.

$-4y-x+x=4x+5+x$ $-4y=5x+5$

Divide both sides of the equation by -4.

$\frac{-4y}{-4}=\frac{5x}{-4}+\frac{5}{-4}$ $y=-\frac{5}{4}x-\frac{5}{4}$

Therefore, the slope of the second equation is -5/4.

Since the slope of each equation is the negative reciprocal of each other, then the graph of the two equations is perpendicular to each other.

5 0

1 year ago

5. Describe the pattern below.<br> 9, 12, 15, 18, 21, 24

Dafna11 [192]

Answer:

The numbers increase by 3

Step-by-step explanation:

The difference between all the numbers in the sequence have a gap of 3

9 and 12 --gap of 3

12 and 15--gap of 3

18 and 21--gap of 3

21 and 24--gap of 3

6 0

2 years ago

Read 2 more answers

sophia earned $52 at her job when she workd for 2 hours. what was her hourly pay rate in hours per dollar?

statuscvo [17]

1 hour =$26, she has an hourly rate of 26 dollars

3 0

3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago

How many solutions are there to the system of equations? StartLayout enlarged left-brace 1st row 4 x minus 5 y = 5 2nd row negat

kkurt [141]

Answer:

Only one solution $x=\frac{5}{4},y=0$

Step-by-step explanation:

$4x-5y=5..............(1)\\0.08x+0.10y=0.10...........(2)$

Multiply equation $(2)$ by $50$

$4x-5y=5........(3)$

Now from equation $(1)$ $+$ equation $(3)$

$8x=10\\x=\frac{10}{8}\\x=\frac{5}{4}$

Put the value in equation $(1)$

$5-5y=5\\5y=0\\y=0$

hence only one solution exists $\left(\frac{5}{4},0\right)$ .

3 0

2 years ago

Read 2 more answers