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3 years ago

Why Is 3 times 1/10 less than 3 times 10?k

Mathematics

2 answers:

vaieri [72.5K]3 years ago

7 0

Answer:

See Explanation

Step-by-step explanation:

When you multiply 3 * 1/10, 3 goes in the numerator so you get 3/10 which is 0.3

On the other hand 3*10 is 30.

0.3 is less than 30 that's why.

Here's the thing you had one pizza slice (equal to 1/10), and your parents gave you three times more the pizza slice. You will have three slices of pizza (each slice being 1/10) not 30 pizzas.

Hope that makes sense. I can explain again if you need me to!

Gennadij [26K]3 years ago

3 0

3 times 1/10 is equal to 3/10, whereas 3 times 10 is equal to 30.

3/ 10 is less than 30.

We can write this as 3 * 1/10 compared to 3 * 10

If we divide both sides by 3, we then compare 1/10 and 10.

1/10 is less than 10, so 3 times 1/10 is less than 3 times 10.

We can rewrite 1/10 = 0.1, so 3 * 0.1 is 0.3 whereas 3 * 10 = 30.

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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 $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .

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3 years ago

Work out the area of a rectangle with base, b = 32mm and perimeter, P = 82mm.

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1 year ago

Read 2 more answers

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3

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