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icang [17]
3 years ago
12

A triangle has sides with lengths of 2x-7, 5x-3, and 2x-2. what is the perimeter of the triangle

Mathematics
2 answers:
Irina18 [472]3 years ago
8 0
Perimeter = 2x - 7 + 5x - 3 + 2x - 2
 = 9x - 12
adell [148]3 years ago
3 0
Perimeter of a triangle =sum of all sides
                                     =2x-7+5x-3+2x-2
                                     =2x+5x+2x-7-3-2
                                     =9x-12 

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Answer:

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3 years ago
What is the area of the trapezoid with height 13 units? Enter your answer in the box. units2 A rectangle with a length of 15 and
jolli1 [7]

Answer:

Step-by-step explanation:

Area of the trapezoid = area of rectangle + 2 * area of triangle

Area of a triangle = 1/2 bh

= 1/2 * 13 * 7

= 45.5 units²

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= 15 × 13

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3 0
3 years ago
Read 2 more answers
Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
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d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
5 0
3 years ago
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Paraphin [41]
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5 0
3 years ago
32% of 300 is what number?
Wewaii [24]

Answer:

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Four options to solve:

1.  32% of 300 = 96

<em>OR</em>

2.  32% × 300 = 96

<em>OR</em>

3.  0.32 × 300 = 96

<em>OR</em>

4.  \frac{32}{100} × 300 = 96

8 0
3 years ago
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