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Zinaida [17]
2 years ago
10

For number 9 and 10 what do I put for the distance between the points?

Mathematics
1 answer:
olga55 [171]2 years ago
4 0

Answer:

I have wrote just answers

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the perimeter of an isosceles triangle in which the equal sides measures 12cm each and third side is 8cm_____. please show the w
KiRa [710]

Answer:

The perimeter of the isosceles triangle is 32 centimeters

Step-by-step explanation:

<em>The perimeter of any figure is </em><em>the sum of the lengths of outline sides</em>

Let us use this fact to solve our question

∵ The perimeter of the triangle is the sum of the lengths of its 3 sides

∵ The triangle is an isosceles triangle

∵ The length of each two equal sides is 12 centimeters

∵ The length of the third side is 8 centimeters

→ Add the lengths of the 3 sides

∴ The perimeter of the triangle = 12 + 12 + 8

∴ The perimeter of the triangle = 32 centimeters

∴ The perimeter of the isosceles triangle is 32 centimeters

8 0
3 years ago
Consider the diagram shown where a ladder is leaning against the side of a building. the base of the ladder is 12ft from the bui
OLga [1]
If the length of the ladder is x then 12/x=cos70, so x=12/cos70=35’ approx, answer c.
7 0
3 years ago
Read 2 more answers
If f is a function that f(f(x)) = 2x² + 1, which is the value of f(f(f(f(3)))? Please help!
Sloan [31]

f(f(3))=2\cdot3^2+1=19\\f(f(f(f(3))))=2\cdot19^2+1=723

5 0
3 years ago
Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
7 0
3 years ago
(r + 14x + 55x + 48) \(x + 6)
balu736 [363]

Answer:

r + 69x + 48 / x + 6

Step-by-step explanation:

3 0
3 years ago
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