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3 years ago

Which ordered pairs are solutions to the equation?

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

6 0

To solve this you want to plug in your x's and y's to see if they match.

A. 3(11)-4=33-4=29 and 29 is not equal to 5 so A is not a solution

B.3(5)-4=15-4=11=11 3(3)-4=9-4=5 and 5 is not equal to 2 so B is not a solution

C. 3(2)-4= 6-4=2 2 is not equal to 3 so C is not a solution

D is the answer. 3(5)-4=11 and 3(2)-4= 2

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There are 12 students running for three student coun-

anyanavicka [17]

Must be

(c) 12.11.10

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(d) 12

8 0

3 years ago

A. 6<br> b 1/6<br> c 4/16<br> d 1/4

Dmitrij [34]

D. 1/4
Divide 4 and 16 and you will get D

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3 years ago

An air traffic controller is tracking two planes.To start Plane A was at an altitude of 990 meters and plane B was at an altitud

soldi70 [24.7K]

The equation representing when the two planes would be at the same altitude is 9x=486

<h3>In terms of mathematics, what is an equation?</h3>

Equation: A declaration that two expressions with variables or integers are equal. In essence, equations are questions, and attempts to systematically identify the solutions to these questions have been the driving forces behind the creation of mathematics.

From (a) we get that: Altitude of Plane A = 990-18x

Altitude of Plane B = 1476-27x

make 890 - 18x = 1476-27x

-18x+27 x = 1476-890

9x=486

X= 54

so pass 54 seconds will the two plane at the same altitude.

To learn more about equations from given link

brainly.com/question/25678139

#SPJ9

8 0

1 year ago

Colleen recorded the type of trees in the woods near her house. She counted 20 spruce trees and 50 maple trees. What is the expe

tankabanditka [31]

Answer:

The answer is A $\frac{2}{7}$

Step-by-step explanation:

3 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