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3 years ago

How do you know whether two quantities x and y are proportional?

Mathematics

2 answers:

ANEK [815]3 years ago

8 0

Answer:

I'm pretty sure it is (a)

Allushta [10]3 years ago

6 0

Answer:

I’m sure it’s C it makes sense I had this question also

Step-by-step explanation:

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Morgarella [4.7K]

Answer:

37.7 inches

Step-by-step explanation:

V=πr^2h

Is the formula for the volume of a cylinder

^2 is exponent 2

r is the radius (2) and h is the height (3)

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AlekseyPX

Answer:

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Simplify the product of

tankabanditka [31]

Answer:

Step-by-step explanation:

$(3+\sqrt{7} )(3-\sqrt{7} )\\\\Now we Can Use Identity (a^{2} -b^{2} =(a+b)(a-b)\\a=3 \\ b=\sqrt{7} \\\\3^{2}-\sqrt{7} ^{2} \\9-7=2$

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8 0

3 years ago

1. Is it proportional?

astra-53 [7]

Answer:

Y=5(x) Yes it is porportional.

Step-by-step explanation:

So lets look at the graph.

We first have 1x = 5y

Well, that makes it seem like for each x there are 5 y.

Lets see the other values:

2x = 10y

Divide both sides by 2 tog et x alone:

1x=5y

Again, for each x, there are 5y.

This is true for each one:

3x=15y

1x=5y

4x=20y

1x=5y

5x=25y

1x=5y

and finally:

6x=30y

1x=5y.

So yes, this is poroportional.

It can be shown as x=5y, since for a single x, there are 5 y's.

Hope this helps!

5 0

3 years ago

plain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = x2 − 2x x2 − 4 if x ≠ 2 1 if x =

ElenaW [278]

Answer with Step-by-step explanation:

We are given that

$f(x)=\left\{\begin{matrix}\dfrac{x^2-2x}{x^2-4}&,if\ \ x\neq 2 \\ 1&,if\ \ x=2\end{matrix}\right.$

We have to explain that why the function is discontinuous at x=2

We know that if function is continuous at x=a then LHL=RHL=f(a).

$f(x)=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}$

LHL=Left hand limit when x <2

Substitute x=2-h

where h is small positive value >0

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2-h}{2-h+2}$

$\lim_{h\rightarrow 0}\frac{2-h}{4-h}=\frac{2}{4}=\frac{1}{2}$

Right hand limit =RHL when x> 2

Substitute

x=2+h

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2+h}{2+h+2}=\lim_{h\rightarrow 0}\frac{2+h}{4+h}$

$=\frac{2}{4}=\frac{1}{2}$

LHL=RHL= $\frac{1}{2}$

f(2)=1

$LHL=RHL\neq f(2)$

Hence, function is discontinuous at x=2

4 0

4 years ago