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ivolga24 [154]
3 years ago
15

You are trying to find the distance from Saint Augustine to Jacksonville. What would be the most appropriate unit of measure to

use?

Mathematics
1 answer:
Hunter-Best [27]3 years ago
5 0

Answer:

it is kilometers

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Can someone solve this 28p³+49b²-16b-28
jok3333 [9.3K]
Here is your answer

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6 0
3 years ago
Algebra 1. please help me. i’m literally struggling , i’ll give an extra 50 points if it’s right and brainliest. - question in p
Arada [10]

Answer:

15x² - 3x - 7

Step-by-step explanation:

(12x² + 2x) - (-3x²+ 5x + 7)

First open the parentheses by applying the distributive property.

To do this, multiply each term you have inside (-3x⅔ + 5x + 7) by -1.

Thus, you would have the following:

12x² + 2x + 3x² - 5x - 7

Add like terms

12x² + 3x² + 2x - 5x - 7

15x² - 3x - 7

6 0
2 years ago
What kind of problem is this?
strojnjashka [21]
4x - 7 < 17
    + 7   + 7
     4x < 24
      4      4
       x < 6

Solution Set: (6, ∞) and {x|x < 6}
8 0
3 years ago
5x – 24 &gt; 26 <br> what is the solution to the inequality?
ozzi

Answer:

x > 10

Step-by-step explanation:

5x - 24 > 26

Add 24 on both sides.

5x -24 + 24 > 26 + 24

5x > 50

Divide by 5 on both sides.

x > 50 ÷ 5

Find x.

x > 10

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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