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3 years ago

The picturre explains the question

Mathematics

2 answers:

Alina [70]3 years ago

3 0

Answer:

x = 47°

Step-by-step explanation:

Since the two angles on a line add up to 180 degrees:

x = 180-133

x = 47°

Ne4ueva [31]3 years ago

3 0

HOPE THIS WILL HELP U..

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PLEAZE HELP IM FAILING THIS CLASS 9) The volume of a box is represented by x' +11x² + 20x – 32. The width of the box is x – 1, a

In-s [12.5K]

Answer:

Step-by-step explanation:

Volume of the box = x³ +11x² + 20x – 32 I think the ' is a typo for ³

the width is x-1 and the height is x+8

Find an expression for the length

Vol = LWH solve L

Vol / (WH) = L so

L = (x³ +11x² + 20x – 32) / (x-1) (x+8)

so it would help to factor the numerator

(x³ +11x² + 20x – 32) I'm willing to bet (x-1) and (x+8) are factors

but I will plot the equation to find the three roots

(x³ +11x² + 20x – 32) = (x-1) (x+8) (x+4)

L = (x³ +11x² + 20x – 32) / (x-1) (x+8)

= (x-1) (x+8) (x+4) / (x-1) (x+8) the (x-1) and (x+8) cancel out leaving

L = (x + 4)

7 0

3 years ago

What is the 9th term in the addition pattern with formula 7n – 3?

Olegator [25]

Replace n with 9 :
7 × 9 - 3 = 60
the answer is C.60

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