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Korvikt [17]
3 years ago
12

( 2-Part Question.)

Mathematics
2 answers:
swat323 years ago
8 0
3x=40.20.... Each book costed $13.40
lys-0071 [83]3 years ago
7 0
B= $40.20/3
The cost of one book is $13.40
You might be interested in
Are triangles PRU and STQ congruent? Question 2 options:
denis23 [38]

PRU and STQ are not congruent because they aren’t the same size.

No, because they aren’t the same size.

<u>Step-by-step explanation:</u>

Both PRU and STQ triangles aren't in the same size, So it is not congruent. Triangles are congruent if two pairs of corresponding angles and a couple of inverse sides are equivalent in the two triangles.

If there are two sets of corresponding angles and a couple of comparing inverse sides that are not equal in measure, at that point the triangles are not congruent.

3 0
3 years ago
A box contains 10 balls, of which 3 are red, 2 are yellow, and 5 are blue. Five balls are randomly selected with replacement. Ca
VLD [36.1K]

Answer:

The required probability is 0.1.

Step-by-step explanation:

red balls = 3

yellow balls =  2

blue balls = 5

Selected balls = 5

Number of elemnets in sample space = 10 C 5 = 1260

Ways to choose 1 red ball and 4 other colours =  (3 C 1 ) x (7 C 4) = 105

Ways to choose 5 balls of other colours = 7 C 5 = 21

So, the probability is

\frac{105}{1260} + \frac {21}{1260}\\\\\frac{126}{1260}=0.1  

8 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
What's the answer to number 3
Evgesh-ka [11]
Do you know what you need to do to solve this problem?
3 0
3 years ago
Please help me it’s due
Kaylis [27]

Answer:

the answer is 22.36 or 22.4

Step-by-step explanation:

7 0
3 years ago
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