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mixer [17]
3 years ago
6

Does a kilogram have more or less mass than a gram?

Mathematics
1 answer:
maks197457 [2]3 years ago
6 0
More, a kilogram is 1000x the mass of a gram
You might be interested in
Determine the x and y intercepts for the line 3x – 5y + 15 = 0. Show your work.
iragen [17]

Answer:

Step-by-step explanation:

The x and y intercepts occur when either x or y = 0

For the y intercept, x = 0

3(0) - 5y + 15 = 0

- 5y + 15 = 0                  Subtract 15 from both sides.

-5y = - 15                       Divide by - 5

-5y / -5 = - 15/-5

y = 3

For x intercept, y = 0

3x - 5(0) + 15 = 0

3x + 15 = 0                    Subtract 15 from both sides

3x = - 15                        Divide by 3

3x/3 = - 15/3

x = - 5

xintercept = (-5,0)

yintercept = (0,3)

3 0
2 years ago
Y=-x-3
nadya68 [22]

Answer: Heyaa! :)

  • <em>1. y=-x-3 = -1</em>

<em>Slope:  </em>−1

<em>y-intercept: </em>(0,−3)

  • <em>2. y=4x-5 =</em> 4

<em>Slope:</em> 4

<em>y-intercept: </em>(0,−5)

  • <em>3. 2x-y=-2 =</em> 2

<em>Slope:</em> 2

<em>y-intercept:</em> (0,2)

  • <em>4. x+y=4 =</em> 1

Slope: −1

<em>y-intercept:</em> (0,4)

  • <em>5. 3x+4y=-12 = - 3/4</em>

<em>Slope: </em>−3/4

<em>y-intercept: </em>

(0,−3)

Hopefully this helps<em> you!</em>

<em />

- Matthew

4 0
2 years ago
How do i multiply fractions
Hatshy [7]
Best Answer:  <span> to multiply fractions you simply take the denominator of both fractions and multiply them, do the same to the numerator, and you have your answer.
ex.
2 5 10
-- * -- = ----
4 8 32

then, of course, you would need to simplify.

so the answer is

10(/2) 5
-------- = -----
32(/2) 16


To multiply mixed numbers, you need to convert them to improper fractions, multiply and then convert them back to a mixed number. For example:
2 1/2 x 5 1/3 = 5/2 x 16 /3

numerator 5 x 16 = 80 denominator 2 x 3 = 6

so now you convert 80/6 into a mixed number

80 / 6 = 13 and 2/6 or, reduced, 13 1/3. </span>
3 0
3 years ago
Read 2 more answers
Describe how to transform the graph of g(x)= ln x into the graph of f(x)= ln (3-x) -2.
Strike441 [17]

Answer:

The graph of g(x) = ㏑x translated 3 units to the right and then reflected

about the y-axis and then translated 2 units down to form the graph of

f(x) = ㏑(3 - x) - 2

Step-by-step explanation:

* Lets talk about the transformation

- If the function f(x) reflected across the x-axis, then the new

 function g(x) = - f(x)

- If the function f(x) reflected across the y-axis, then the new

 function g(x) = f(-x)

- If the function f(x) translated horizontally to the right  

 by h units, then the new function g(x) = f(x - h)

- If the function f(x) translated horizontally to the left  

 by h units, then the new function g(x) = f(x + h)

- If the function f(x) translated vertically up  

 by k units, then the new function g(x) = f(x) + k

- If the function f(x) translated vertically down  

 by k units, then the new function g(x) = f(x) – k

* lets solve the problem

∵ Graph of g(x) = ㏑x is transformed into graph of f(x) = ㏑(3 - x) - 2

- ㏑x becomes ㏑(3 - x)

∵ ㏑(3 - x) = ㏑(-x + 3)

- Take (-) as a common factor

∴ ㏑(-x + 3) = ㏑[-(x - 3)]

∵ x changed to x - 3

∴ The function g(x) translated 3 units to the right

∵ There is (-) out the bracket (x - 3) that means we change the sign

  of x then we will reflect the function about the y-axis

∴ g(x) translated 3 units to the right and then reflected about the

   y-axis

∵ g(x) changed to f(x) = ㏑(3 - x) - 2

∵ We subtract 2 from g(x) after horizontal translation and reflection

  about y-axis

∴ We translate g(x) 2 units down

∴ g(x) translated 3 units to the right and then reflected about the

   y-axis and then translated 2 units down

* The graph of g(x) = ㏑x translated 3 units to the right and then

  reflected about the y-axis and then translated 2 units down to

  form the graph of f(x) = ㏑(3 - x) - 2

7 0
3 years ago
Read 2 more answers
(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t&lt;00≤t&lt;11≤t&lt;7t≥7;f(t)={0,t&lt;0−5,0≤t&lt;1−6,1≤t&lt;76,t≥7; 1. wr
sergij07 [2.7K]
f(t)=\begin{cases}0&\text{for }t

Recall that

u(t)=\begin{cases}0&\text{for }t

Take it one piece at a time. For t\ge0, we can scale u(t) by -5:

-5u(t)=\begin{cases}0&\text{for }t

If we shift the argument by 1 and scale by -5, we have

-5u(t-1)=\begin{cases}0&\text{for }t

so if we subtract this from -5u(t), we'll end up with

-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t

For the next piece, we can add another scaled and shifted step like

-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

so that

-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

For the last piece, we add one more term:

6u(t-7)=\begin{cases}0&\text{for }t

and so putting everything together, we get f(t):

f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)
f(t)\equiv-5u(t)-u(t-1)+12u(t-7)
5 0
3 years ago
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