Bria had 6 yards of fabric and she used 2 3/4 yards on a skirt. How much fabric is left?

Mathematics

2 answers:

solong [7]3 years ago

8 0

Answer: 3 1/4 yards

Step-by-step explanation:

6 yards - 2 3/4 = 3 1/4

kotegsom [21]3 years ago

3 0

Answer:

3 ¹/₄ yards of fabric left

Step-by-step explanation:

First, we need to set up our subtraction equation as shown:

6 - 2 ³/₄ =

Next, we are going to convert both terms into improper fractions by multiplying the whole number by the decimal and then adding the numerators.

6 x 4 = 24 so 6 becomes ²⁴/₄

2 x 4 = 8 so 2 becomes ⁸/₄ plus the ³/₄ we began with giving us a total of ¹¹/₄

Then, subtract the numerators and simply if possible.

²⁴/₄ - ¹¹/₄ = ¹³/₄

¹³/₄ = 3 ¹/₄ yards of fabric left

~Hope this Helped!~

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Step-by-step explanation:

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