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tatiyna
2 years ago
5

Plz help find area and perimeter

Mathematics
1 answer:
kati45 [8]2 years ago
3 0

Answer:

okay

Step-by-step explanation:

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HELP PLEASE NEED TO PASS!!!!!!!!!Rob is investigating the effects of font size on the number of words that fit on a page. He cha
zepelin [54]

Using the given data you can build the linear regression (described by linear function y=ax+b) by hand or you can use math calculator to do this. In attached diagram you can see that

a=-26.059\approx -26

and

b=722.633\approx 723.

Then the equation of linear function is

y = -26x + 723.

Answer: correct choice is D.

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3 years ago
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Triangle DIG is reflected across the l-axis.
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there is not such thing as an I axis, but I will show you what happens for both x and y axis so you can fill out the assignment with the correct information.

IF you meant flipping over the x axis, then the dot will go from top to bottom, bottom to up, the x position won't change only its height(y position)

This meant that for reflecting across the x axis, (x,y)->(x,-y)

For y axis its the opposite, just flipping to the side.

This meant that for reflecting across the y axis, (x,y)->(-x,y)

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2 years ago
What percent of 45 is 135? Explain your reasoning.
MrRissso [65]
The question is equal to: 135 / 45

Divide:

135 / 45 = 3

3 = 300%

Answer = 300% (I think)

~Hope this helped~


6 0
3 years ago
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The models below are shared to show 4/8 and 2/4. What do the models show about these fractions
Vanyuwa [196]

Answer:

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3 years ago
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The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasin
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Answer:

\frac{dI}{dt}=-0.00016 A/s

Step-by-step explanation:

We are given that

By ohm's law

V=IR

R=389 ohm

I=0.03 A

\frac{dV}{dt}=-0.06V/s

\frac{dR}{dt}=0.07ohm/s

We have to find rate of change of current I means \frac{dI}{dt}

Differentiate the equation w.r.t t

\frac{dV}{dt}=\frac{dI}{dt}R+I\frac{dR}{dt}

Substitute the values then we get

-0.06=\frac{dI}{dt}\times 389+0.03\times 0.07

-0.06=389\frac{dI}{dt}+0.0021

-0.06-0.0021=389\frac{dI}{dt}

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\frac{dI}{dt}=\frac{-0.0621}{389}=-0.000160 A/s

Hence, the current I is changing at the rate=-0.00016A/s

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