( 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

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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 $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .