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3 years ago

Help me please and thank you

Mathematics

1 answer:

erica [24]3 years ago

7 0

Answer:

ΔMXP, ΔXMP, ΔPMX

Step-by-step explanation:

A triangle has three sides and three points, so it can only be named using the three points, X, P and M.

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What is 61 divided by 207.4

svet-max [94.6K]

Answer:

5/17 or 0.294...

Step-by-step explanation:

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6 0

3 years ago

4) If the area of a triangle is

sineoko [7]

Answer:

The height is 12 feet.

Step-by-step explanation:

The formula for area of a triangle is 1/2 ⋅ b (base) ⋅ h (height).

Now let's plug in the numbers.

According to the word problem, the base is 12 feet.

As of now, our equation is: 1/2 ⋅ 12 ⋅ h

The only number times 1/2 that would give you an answer of 72 is 144.

12⋅12 = 144.

This means the height is 12.

And, 1/2 ⋅ 12 ⋅ 12 ⋅ = 72.

Because the area of the triangle is 72, 72 is the number we are looking for.

Therefore, the height is 12.

I really hope this made sense, it was a bit rushed.

6 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