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3 years ago

A wooden board in the shape of a rectangular prism measures 0.3 m by 2.1 m by 0.1 m and has a mass of 0.07 kg.

1 answer:

6 0

Density = Mass/Volume
D = M/V
D = 0.07/0.063
D = 1.111111111111111

Volume of a rectangular prism is WHL
Volume of a rectangular prism is 0.3 m * 2.1 m * 0.1 m
Volume of a rectangular prism is 0.063

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HELP ME PLEASE I WILL GIVE BRAINLIEST FOR BOTH PARTS PLEASE HELP

Katen [24]

For part A: two transformations will be used. First we will translate ABCD down 3 units: or the notation version for all (x,y) → (x, y - 3) so our new coordinates of ABCD will be:
A(-4,1)
B(-2,-1)
C(-2,-4)
D(-4,-2)

The second transformation will be to reflect across the 'y' axis. Or, the specific notation would be: for all (x,y) → (-x, y) New coordinates for A'B'C'D'
A'(4,1)
B'(2,-1)
C'(2,-4)
D'(4,-2)

Part B: The two figures are congruent.. We can see this a couple of different ways.
- first after performing the two transformations above, you will see that the original figure perfectly fits on top of the image.. exactly the same shape and size.
- alternatively, you can see that the original and image are both parallelograms with the same dimensions.

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3 years ago

*Asymptotes* g(x) =2x+1/x-3 Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

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3 years ago

Graph the line giving one point and the slope. 2x-6y=12

koban [17]

Y = mx + c

this is the equation of line and here m is the slope

so, 2x-6y =12 can be changed into standard format

y = 2x/6 - 12/6
y=x/3 +(-2)

so, m = 1/3 , so slope will be 1/3

hope it helped :)

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3 years ago

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Khan sucks 3 for 30 points again

tester [92]

Answer:

D, John did not make a mistake!

Step-by-step explanation:

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2 years ago

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