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mestny [16]
3 years ago
13

Use the graph of the function to find its average rate of change from

Mathematics
1 answer:
olasank [31]3 years ago
7 0
 <span>Use the graph of the function to find its average rate of change from 
=x−3
to 
=x3 </span>is x<2\3
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What is next step in solving this equation <br> 32x+2=40
Alenkinab [10]

Answer:

32x+2=40

-2 -2

32x=42

.................

5 0
3 years ago
Find the slope of the line that passes through the points (1, 1) and (7, 5).
Nata [24]

Step-by-step explanation:

m=y²-y¹/x²-x¹

m=5-1/7-1

m=4/6

m= 2/3

The slope is 2/3

7 0
2 years ago
Read 2 more answers
The 10 students who completed a special flying course are waiting to see if they will be awarded the one Distinction or the one
yaroslaw [1]

Answer: 1)100ways and 2)90ways


Step-by-step explanation:

Given :-The 10 students who completed a special flying course are waiting to see if they will be awarded the one Distinction or the one Merit award for their efforts.

Ways can the two awards be given if the same student can receive both awards (i.e repetition is allowed)

=10×10=100ways

Ways can the two awards be given if the same student cannot receive both awards(i.e. repetition is not allowed)

=10×9=90ways

3 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
A right triangular prism is shown.What is the lateral surface area?
makkiz [27]

The general formula for the lateral surface area of a right prism is

L = ph

where p represents the perimeter of the base and h represents the height of the prism.

In our case, the perimeter is the sum of the bases.

p = 5 + 5.6 + 2.5 = 13.1 cm

now, the height of the prism is 8 cm

then L = 13.1 cm x 8 cm = 104.8cm^2

5 0
9 months ago
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