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photoshop1234 [79]
2 years ago
5

During batting practice, two pop flies are hit from the same location, 2 s apart. The paths are modeled by the equations h = -16

t^2 + 56t and h = -16t^2 + 156t - 248, where t is the time that has passed since the first ball was hit. Explain how to find the height at which the balls meet. Then find the height to the nearest tenth.
Mathematics
1 answer:
Schach [20]2 years ago
4 0

Answer:

2.5 seconds since the first ball was hit.

Step-by-step explanation:

The equations that model the heights are:

  • h = -16t² + 56t
  • h = -16t² + 156t - 248

When the balls meet, both heights are equal, therefore:

-16t² + 56t = -16t² + 156t - 248

-16t² + 56t + 16t² - 156t + 248 = 0

- 100t + 248 = 0

248 = 100t

248/100 = t

t = 2.48 ≈ 2.5

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Yuri is thinking of a 4-digit whole number. He rounds his number to the nearest thousand. His answer is 4000, what is the smalle
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Answer:

Smallest number = 3500

Step-by-step explanation:

Rounding of numbers involve replacing numbers with simpler numbers. In order to round a number to the nearest thousand, the last 3 digits of the number should be considered. If the last 3 digits are less than 500, the number is rounded down(the thousand figure is unaffected), but if the last 3 digits are greater or equal to 500, the number is rounded up.

In this case, Yuri is thinking of a 4-digit whole number and he rounds his number to the nearest thousand. Since his answer is 4000, the smallest number yuri could be thinking of would be 3500 and the highest number he could be thinking of is 4499.

Thus, the smallest number Yuri could be thinking of is 3500

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Find the equation of the plane through the point (2,5,7) that is parallel to the line r=(3i+2j−2k)+t(i+2j+9k) and perpendicular
insens350 [35]

The plane we want to find has general equation

a(x-2)+b(y-5)+c(z-7)=0

with a,b,c not equal to 0, and has normal vector

\vec n=a\,\vec\imath+b\,\vec\jmath+c\,\vec k

\vec n is perpendicular to both the normal vector of the other plane, which is 4\,\vec\imath+5\,\vec\jmath+6\,\vec k, as well as the tangent vector to the line \vec r(t), which is \vec\imath+2\,\vec\jmath+9\,\vec k.

This means the dot product of \vec n with either vector is 0, giving us

\begin{cases}4a+5b+6c=0\\a+2b+9c=0\end{cases}

Suppose we fix c=1. Then the system reduces to

\begin{cases}4a+5b=-6\\a+2b=-9\end{cases}

and we get

(4a+5b)-4(a+2b)=-6-4(-9)\implies-3b=30\implies b=-10

a+2(-10)=-9\implies a=11

Then one equation for the plane could be

\boxed{11(x-2)-10(y-5)+(z-7)=0}

or in standard form,

\boxed{11x-10y+z=-21}

The solution is unique up to non-zero scalar multiplication, which is to say that any equation (11x-10y+z)k=-21k would be a valid answer. For example, suppose we instead let c=2; then we would have found a=22 and b=-20, but clearly dividing both sides of the equation

22(x-2)-20(y-5)+2(z-7)=0

by 2 gives the same equation as before.

7 0
2 years ago
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