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aev [14]
2 years ago
10

Which side is adjacent to

Mathematics
1 answer:
nexus9112 [7]2 years ago
7 0

Answer:

from B-A is adjacent

because of the 90 degree

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Solve for x I need help on questions 5 & 6
Effectus [21]

Answer:

#5 x=8, #6 x=5

Step-by-step explanation:

Theyre vertical so for number 5 you would have the equation

5x+15=55

    -15   -15 from both sides

5x=40

divide by 5 from both sides and you get

x=8

Same thing vertical so you would have  

16x+7=87

    -7    -7 from both sides

16x=80

divide by 16 from both sides

x=5

3 0
2 years ago
Find g (x) - f (x). <br><br> g (x) = x^2 + 1 and f (x) = 2x + 5
allochka39001 [22]

Answer:

x^2 - 2x - 4

Step-by-step explanation:

g(x) - f(x) = (x^2 + 1) - (2x + 5) = x^2 + 1 - 2x -5 = x^2 -2x +(1-5) = x^2 - 2x - 4

6 0
3 years ago
Quadrilateral ABCD is reflected across the y-axis and translated 1 unit down to create quadrilateral EFGH .
sleet_krkn [62]
GH is parallel to FE
8 0
3 years ago
Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
d^*(2,10,2\sqrt5)=46
d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
5 0
3 years ago
Which of the following is the appropriate interpretation for when an event has a probability of 1? Group of answer choices We ar
ira [324]

9514 1404 393

Answer:

  We are absolutely certain that the event will occur

Step-by-step explanation:

An event with probability 1 is certain to occur. There is no chance it will not occur.

We are absolutely certain that the event will occur.

6 0
3 years ago
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