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3 years ago

Plz help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Mathematics

1 answer:

iren2701 [21]3 years ago

7 0

Answer:

88 inches

Step-by-step explanation:

The given bicycle has a tyre that is 28 inches in diameter.

How far the bicycle moves forward each time the wheel goes around is the circumference of the bicycle tyre.

This is calculated using the formula:

C =\pi \: d

We substitute the diameter and

\pi = \frac{22}{7}

C = \frac{22}{7} \times 28

This simplifies to

C =22 \times 4

88 \: inches

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Radical expressions Help pls multiple choice

Ann [662]

Answer:

4th option (sqrt(10))/2

Step-by-step explanation:

$\frac{5}{\sqrt{10} } * \frac{\sqrt{10}}{\sqrt{10} } \\ \frac{5\sqrt{10}}{10} \\\\frac{\sqrt{10}}{2}$

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3 years ago

What is the additive inverse of the complex number -8 + 3i? -8-3i -8+3i 8-3i 8+3i

Nady [450]

The answer is b because the answer is b and the people don’t know how to do it

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A square tile has an area of 324 square inches. what is the perimeter of the square tile in inches?

kotykmax [81]

The formula of an area of a square: $A=a^2$

a - side

We have: $A=324\ in^2$

substitute:

$a^2=324\to a=\sqrt{324} \to a=18\ in$

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substitute:

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7 0

3 years ago

Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−

aleksandrvk [35]

The tangent to $C$ through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces $S$ and $T$ at that point.

Let $f(x,y,z)=x^2+2y^3+3z^4$ . Then $S$ is the level curve $f(x,y,z)=6$ . Recall that the gradient vector is perpendicular to level curves; we have

$\nabla f(x,y,z)=(2x,6y,12z^2)$

so that the gradient of $f$ at (1, 1, 1) is

$\nabla f(1,1,1)=(2,6,12)$

For the surface $T$ , we have

$\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0$

so that $\vec r(1,0)=(1,1,1)$ . We can obtain a vector normal to $T$ by taking the cross product of the partial derivatives of $\vec r(u,v)$ , and evaluating that product for $u=1,v=0$ :

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)$

$\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)$

Now take the cross product of the two normal vectors to $S$ and $T$ :

$(2,6,12)\times(1,-1,2)=(24,8,-8)$

The direction of vector (24, 8, -8) is the direction of the tangent line to $C$ at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by $t\in\Bbb R$ . Then adding (1, 1, 1) shifts this line to the point of tangency on $C$ . So the tangent line has equation

$\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)$

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3 years ago

Write -a^3b^6 and -27x^6b^9 as cubes of a monomial. -a^3b^6=( ) ^3 -27x^6b^9=( )^3

Brrunno [24]

Answer:

I'm sorry I don't know the first one but the second one is (-3x^2b^3)^3

Step-by-step explanation:

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3 years ago