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vodka [1.7K]
3 years ago
6

Where do you get the .95 from?

Mathematics
1 answer:
yan [13]3 years ago
4 0

Answer:

by subtracting .05 from 1

Step-by-step explanation:

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Write your own real-world scenario where the Pythagorean Theorem can be applied to find a missing piece. You may choose to write
Nookie1986 [14]

Answer:

Bucky's ship is located 43 miles west off the harbor. Clyde's yacht is located 25 miles north from Bucky. How far is Clyde from the Harbor?

The picture would explain how it relates to the Pythagorean Theorem.

Brainilest is appreciated.

7 0
3 years ago
What numbers go in the blank!?!
Inessa [10]
16 is the correct anwser
7 0
3 years ago
The radii of two circles are in the ratio of 3 to 1. Find the area of the smaller circle if the area of the larger circle is 27
Naddik [55]
ANSWER
3\pi sq.\: in.

EXPLANATION

Let R be the radius of the bigger circle and r, be the radius of the smaller circle.

Then their ratio is given as,

R:r=3:1

We can rewrite it as fractions to get,

\frac{R}{r} = \frac{3}{1}

We make R the subject to get,

R = 3r

The area of the bigger circle can be found using the formula,

Area=\pi {r}^{2}

This implies that,

Area=\pi ({3r})^{2}

Area=9\pi {r}^{2}

But it was given in the question that, the area of the bigger circle is 27π.

27\pi=9\pi {r}^{2}

We divide through by 9π to get,

3 = {r}^{2}

This means that,
r = \sqrt{3}

The area of the smaller circle is therefore

= \pi {( \sqrt{3}) }^{2}

= 3\pi
8 0
4 years ago
Read 2 more answers
A soccer goalie kicks the soccer ball. The quadratic function y = −16t2 + 64t gives the time t seconds when the soccer ball is a
zzz [600]

Answer:

The correct answer is 4 sec

Step-by-step explanation:

Download docx
7 0
3 years ago
find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2
Alexxandr [17]

The surface is parameterized by

\vec s(u,v) = x(u, v) \, \vec\imath + y(u, v) \, \vec\jmath + z(u, v)) \, \vec k

and the normal to the surface is given by the cross product of the partial derivatives of \vec s :

\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}

It looks like you're given

\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}

Then the normal vector is

\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k

Now, the point (2, 2, 0) corresponds to u and v such that

\begin{cases}u + v = 2\\2u^2 = 2\\u - v = 0\end{cases}

and solving gives u = v = 1, so the normal vector at the point we care about is

\vec n = -4\,\vec\imath+2\,\vec\jmath-4\,\vec k

Then the equation of the tangent plane is

\left(-4\,\vec\imath + 2\,\vec\jmath - 4\,\vec k\right) \cdot \left((x-2)\,\vec\imath + (y-2)\,\vec\jmath +  (z-0)\,\vec k\right) = 0

-4(x-2) + 2(y-2) - 4z = 0

\boxed{2x - y + 2z = 2}

6 0
2 years ago
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