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3 years ago

Look at the picture for the question Please answer

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

7 0

Answer:

$\large\boxed{D.\ 12x^2-29x+14}$

Step-by-step explanation:

Use FOIL: <em>(a + b)(c + d) = ac + ad + bc + bd</em>

$(3x-2)(4x-7)=(3x)(4x)+(3x)(-7)+(-2)(4x)+(-2)(-7)\\\\=12x^2-21x-8x+14\qquad\qquad\text{combine like terms}\\\\=12x^2+(-21x-8x)+14=12x^2-29x+14$

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His brothers have 4 games
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So in total there are 14 games
And if there are 2 cases and each one had the same amount, then in each one there would be 14÷2=7

In each case, there are 7 games.

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3 years ago

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Answer:

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Step-by-step explanation:

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Explanation

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3 years ago

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3 years ago

Read 2 more answers

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$

d) ON the same fashion we want to know the norm of the second derivative of $\gamma$ .

We have that $\gamma ^{''}(t) =(-10\cost t, -10 \sin t , 0)$ so the acceleration is given by $\sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10$

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3 years ago