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3 years ago

10

This graph shows a proportional relationship. What is the constant of proportionality?

2 answers:

krek1111 [17]3 years ago

7 0

hookes law gives the equation F=kx where F is the elastic force and k is the constant and x or small e is extension if we draw a graph you'll see that the graph increases by the same ratio every single time hence giving a straight line show that they are F and x are propotional to a certain limit

spayn [35]3 years ago

3 0

Answer:

The answer you're looking for is 40

Step-by-step explanation:

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15) 5,25 , 125 ,625 ,3125<br>Rule =

MrMuchimi

Answer:

multiply by 5

Step-by-step explanation:

5x5=25

25x5+125

125x5=625

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3 years ago

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3. 6076 divided by 98

cluponka [151]

Answer:

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Nookie1986 [14]

Answer:

$\large\boxed{y=\dfrac{1}{4}x^2-x-4}$

Step-by-step explanation:

The equation of a parabola in vertex form:

$y=a(x-h)^2+k$

<em>(h, k)</em><em> - vertex</em>

The focus is

$\left(h,\ k+\dfrac{1}{4a}\right)$

We have the vertex (2, -5) and the focus (2, -4).

Calculate the value of <em>a</em> using $k+\dfrac{1}{4a}$

<em>k = -5</em>

$-5+\dfrac{1}{4a}=-4$ <em>add 5 to both sides</em>

$\dfrac{1}{4a}=1$ <em>multiply both sides by 4</em>

$4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4$

$\dfrac{1}{a}=4\to a=\dfrac{1}{4}$

Substitute

$a=\dfrac{1}{4},\ h=2,\ k=-5$

to the vertex form of an equation of a parabola:

$y=\dfrac{1}{4}(x-2)^2-5$

The standard form:

$y=ax^2+bx+c$

Convert using

$(a-b)^2=a^2-2ab+b^2$

$y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5$

<em>use the distributive property: a(b+c)=ab+ac</em>

$y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4$

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4 years ago

arc GT = 18 degrees, arc TA =32 degrees, arc NA =112 degrees, arc IN = 74 degrees. find measure of angle GET

Svet_ta [14]

Answer:

Step-by-step explanation:

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3 years ago