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Anuta_ua [19.1K]
3 years ago
5

Complete the square and put this function in vertex form: f(x)=x^2+20x+97

Mathematics
1 answer:
Fofino [41]3 years ago
3 0

Answer:

Step-by-step explanation:

f(x)=(x²+20x+100) -3  because : 97 = 100 -  3

f(x) = (x+10)² -3 ....vertex form

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10x^2-10x+20 is the answer to this
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[Calculus] particle at rest question. Show steps, please!
Vinvika [58]

Answer:

E

Step-by-step explanation:

We are given that a particle's position along the x-axis at time <em>t </em>is modeled by:

x(t)=2t^3-21t^2+72t-53

And we want to determine at which time(s) <em>t</em> is the particle at rest.

If the particle is at rest, this suggests that its velocity at that time is 0.

Since are we given the position function, we can differentiate it to find the velocity function.

So, by differentiating both sides with respect to <em>t</em>, we acquire:

\displaystyle x^\prime(t)=v(t)=\frac{d}{dt}\big[2t^3-21t^2+72t-53\big]

Differentiate. So, our velocity function is:

v(t)=6t^2-42t+72

So, we will set the velocity to 0 and solve for <em>t</em>. Hence:

0=6t^2-42t+72

We can divide both sides by 6:

0=t^2-7t+12

Factoring yields:

(t-3)(t-4)=0

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t-3\text{ and } t-4=0

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t=3\text{ and } t=4

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6 0
3 years ago
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Answer:

Step-by-step explanation:

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4 0
4 years ago
Saturn is 8.867 × 10^8 miles away from the Sun. Uranus is 1.787 × 10^9 miles away from the Sun. Approximately how many times far
Over [174]

Answer:

<em>2 times </em>

<em></em>

Step-by-step explanation:

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Distance of Uranus from Sun, D_{2} = 1.787 \times 10^{9}\ miles

We have to find, how many times the Uranus is far from Sun than Saturn is far from Sun.

Let us begin by multiplying the distance of Saturn from Sun by 2: D_{1} \times 2 = 8.867 \times 10^{8}\ miles \times 2\\\Rightarrow 17.734 \times 10^{8}\ miles\\\Rightarrow 1.7734 \times 10^{1} \times  10^{8}\ miles\\\Rightarrow 1.7734 \times 10^{8+1}\  miles\\\Rightarrow 1.7734 \times 10^{9} $\approx$ D_{2}

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3 0
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4 years ago
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