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Elodia [21]
3 years ago
6

Which of the following has a graph that is wider than the graph of y=3x2+2 ? a.y=3x2+3 b.y=0.5x2+1 c.y=4x2 +1 d.y=4x2+1

Mathematics
2 answers:
AveGali [126]3 years ago
6 0

Answer:

Option B is correct.

Step-by-step explanation:

Given: y=3x^2+2

We need to choose correct graph whose wider than given graph.

For wider, we check the coefficient of x².

For function y=ax²

If a>1 then compressed

If a<1 then wider

We will compare the coefficient of x² for each option.

Option A: y=3x^2+2

3x^2+2\rightarrow 3x^2+2 Identical graph.

Option B: y=0.5x^2+1

3x^2+2\rightarrow 0.5x^2+1

3>0.5, Graph becomes wider than older one.

Option C: y=3x^2+2

3x^2+2\rightarrow 4x^2+1

3<4, Graph thinner than original graph.  

Option D: y=3x^2+2

3x^2+2\rightarrow 4x^2+1

3<4, Graph is thinner than original graph.

Hence, Option B is correct. (Please see attachment for graph)

Goryan [66]3 years ago
5 0
So sorry fir it being this late but the answer is B
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1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

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\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

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