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faltersainse [42]
3 years ago
6

Problems 20-22 has me mixed up​

Mathematics
1 answer:
drek231 [11]3 years ago
8 0

See the attached picture:

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10. A box contains five and two-thirds cups of rice. If three fourths of the rice will
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Answer:

5 2/3 - 3/4

5 8/12 - 9/12

4 20/12 - 9/12

4 11/12

4 0
3 years ago
If you are given the graph of a line in an xyplane and its corresponding equation, you can be sure that the coordinates of every
topjm [15]

If the graph of a line  in an x y plane and its corresponding equation is given then , all the coordinates of every point on that line will satisfy the equation is a true statement.

<h3>What is the equation for a straight line ?</h3>

An equation of a straight line is given by

y = mx+c

where m is the slope and c is the intercept on y axis.

If the graph of a line  in an x y plane and its corresponding equation is given then , all the coordinates of every point on that line will satisfy the equation .

Even for a scatter plot the line of the best fit is drawn and every point on the line will satisfy the equation.

To know more about Equation of Straight Line

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8 0
2 years ago
5x - 8y = 20<br> what are x and y? :)
Alex_Xolod [135]

Answer:

x could be 10

y could be 5

those are some possible answers.

8 0
2 years ago
Read 2 more answers
Please help me add this fractions(show work) ​
tatuchka [14]

Answer:

Step-by-step explanation:

6 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
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