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Oduvanchick [21]
3 years ago
6

How many students started in ninth grade if the number of graduates is 25,000?

Mathematics
1 answer:
Mama L [17]3 years ago
5 0

Answer:

Are you in Connexus ?

Step-by-step explanation:

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Mrs McCarr buys 9 packages of markers for an art project. Each packages has 10 markers. How many markers does Mrs McCarr buy?
Alborosie

Answer:

Mrs. McCarr buys 90 markers

Step-by-step explanation:

10 markers per package × 9 packages of markers = 90 markers in total

5 0
3 years ago
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Use complete sentences to describe the domain of the cosine function.
Hatshy [7]
The domain of a function is the set of the possible input values of the function. For example: consider the function f(x) = cos x, the domain of the function is the set of possible values of x.

The cosine function takes x values from all real numbers.

Therefore, the domain of the cosine function is a real numbers.
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
Please help ASAP! Will give BRAINLIEST! Please read the question THEN answer correctly! No guessing.
Phantasy [73]
Option C is correct.

try multiplying it out yourself. It should work
7 0
3 years ago
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Help please!!!!..............
Irina18 [472]

Answer:

hi so i think its a

Step-by-step explanation:

4 0
3 years ago
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