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Nezavi [6.7K]
2 years ago
13

Help me ! I need all the help I can get

Mathematics
1 answer:
anygoal [31]2 years ago
7 0

Although i could be wrong i think that the linear equation for the question would be

x=1/5y+2/5

The y intercept for the equation is 1/5

The slope is 2/5

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A circle with radius 4 inches is inscribed in an equilateral triangle. Find the area of the triangle.
LUCKY_DIMON [66]

The inscribed circle has its center at the point of intersection of the angle bisectors, which also happen to be the medians. Hence the altitude of the triangle is 3 times the radius, or 12 inches.

The side length of this triangle is 2/√3 times the altitude, so the area is

... Area = (1/2)·b·h = (1/2)·(24/√3 in)·(12 in)

... Area = 48√3 in² ≈ 83.1384 in²

7 0
3 years ago
Someone help me please
Gemiola [76]

Answer:

x=2\sqrt{6}

Step-by-step explanation:

Use the Pythagorean theorem:

a^2+b^2=c^2

a and b are the legs and c is the hypotenuse. Insert the values:

\sqrt{10}^2+\sqrt{14}^2=x^2

Simplify exponents using the rule \sqrt{x}^2=x :

10+14=x^2

Simplify addition:

24=x^2\\x^2=24

Find the square root:

\sqrt{x^2}=\sqrt{24}\\  x=\sqrt{24}

Simplify in radical form: Find a common factor of 24 that is a perfect square:

\sqrt{24}=\sqrt{4*6}

Separate:

\sqrt{4}*\sqrt{6}

Simplify:

2*\sqrt{6}\\x=2\sqrt{6}

Finito.

3 0
3 years ago
Read 2 more answers
Find an equation of the tangent plane to the given parametric surface at the specified point.
Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

Step-by-step explanation:

Given equation

      r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}---(1)

Normal vector  tangent to plane is:

\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}

\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}

Normal vector  tangent to plane is given by:

r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]

Expanding with first row

\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\

at u=5, v =π/3

                  =\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k} ---(2)

at u=5, v =π/3 (1) becomes,

                 r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

From above eq coordinates of r₀ can be found as:

            r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})

From (2) coordinates of normal vector can be found as

            n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)  

Equation of tangent line can be found as:

  (\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

5 0
3 years ago
If h= mv-l+h than mv+l-h=
kotegsom [21]

Answer:

l

Step-by-step explanation:

3 0
3 years ago
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The scale of a model train is 1 inch to 13.5 feet. One of the cars of the model trial is 5 inches long. What is the length, in f
garik1379 [7]
If 1 inch is 13.5 feet, then 5 inches = (13.5 * 5) = 67.5 ft
8 0
3 years ago
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