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balu736 [363]
3 years ago
12

Does the point (8, 2)satisfy the inequality 15x +18y = 16 YES OR NO

Mathematics
1 answer:
Marrrta [24]3 years ago
5 0

false.No

Step-by-step explanation:

15x+18y=16

(8,2)

15(8)+18(2)=16

120+36=16

156=16

false

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HELP PLEASE! THIS IS MY LAST QUESTION ILL GIVE BRAINLIEST, NO LINKS. <3
Elena L [17]

Answer: C {3, 4}

Step-by-step explanation:

<em>Solve the inequality:</em>

7x + 6 > 20

7x > 20 - 6

7x > 14

x > 2

Only 3 and 4 are greater than 2.

7 0
2 years ago
R+(-5r)<br> I really need a step by step answer
Dennis_Churaev [7]

Answer:

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r-5r

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5 0
3 years ago
x,y, and z to the nearest tenth .(hint: use altitude and leg rules) PLEASE HELP FAST !!! please solve and show steps
madreJ [45]

Use the Pythagorean theorem to find z:

z = √(6^2 + 9^2)

z= √(36 + 81)

z = √117

z = 3√13 = 10.8

Use leg rule to find x:

x/6 = 6/9

Cross multiply:

9x = 36

Divide both sides by 9:

x = 36/9

x = 4

Use Pythagorean theorem to find y:

y = √(6^2 + 4^2)

y= √(36 + 16)

y = √52

y = 2√13 = 7.2

7 0
3 years ago
Phil has two rectangular prisms. The height of the second prism is 1/2 times the height of the first prism. The lengths and widt
Vesnalui [34]

The volume of the second prism is also ten times the volume of the first prism.

Let's assume that both prisms have:

width = 3 units

height = 4 units

Prism 1 length = 5 units

Prism 2 length = 50 units

Let's solve their respective volumes to compare...

Volume of prism 1 = length * width * height

                             = 5 * 3 * 4

                             = 60 units ^3

Volume of prism 2 = 50 * 3 * 4

                             = 600 units ^3

Prism 2/ prism 1 = 10

That means prism 2 is ten times the volume of prism 1.

6 0
2 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
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