Is the ordered pair (1, 4) a solution to the equation y = 2x + 1?If yes, is it the only solution?

A.
No, because the ordered pair does not fit the equation.

B.
Yes, but this ordered pair is not the only solution to the equation.

C.
Yes, and it is the only ordered pair that fits the equation.

Answers

Answer 1

Answer: Let's plug in x-coordinate x=1:

y = 2(1)+1 = 2 + 1 = 3, which is not 4.

So the answer is A.

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suppose that when the price is $10, a deli is willing to sell 200 sandwiches. if the price falls to $8, how many sandwiches would the deli be willing to sell

Answers

Answer:

Step-by-step explanation:

we'll cross-multiply

(200)(8) = 100x

1600/10 = x

160 = x

He will want to sell 160 sandwiches for $8

What two numbers add up to 6 but multiply to 1

Answers

The two numbers (rounded) are

5.84823 and 0.17157

What is 165 as a mixed number

Answers

165 percent is equal to 1 13/20

please leave thanks

Which equation represents a hyperbola with a center at (0, 0), a vertex at (0, 60), and a focus at (0, −65)?

Answers

Answer:

Step-by-step explanation:

on edge

Answer:

D!!

Step-by-step explanation:

Got it right

I need help solving x^2+3xy-y^2=12 and x^2-y^2=-12

Answers

$substitute:x^2-y^2=-12\ to\ x^2+3xy-y^2=12\n\n3xy-12=12\n3xy=12+12\n3xy=24\ \ \ \ \ /:3\nxy=8\to x=(8)/(y)\n\nsubstitute\ to\ x^2-y^2=-12\n\n\left((8)/(y)\right)^2-y^2=-12\n\n(64)/(y^2)=y^2-12$

$(64)/(y^2)=(y^2-12)/(1)\n\ny^2(y^2-12)=64\n\ny^4-12y^2-64=0\n\nsubstitute:y^2=t > 0\n\nt^2-12t-64=0$

$a=1;\ b=-12;\ c=-64\n\Delta=b^2-4ac\to\Delta=(-12)^2-4\cdot1\cdot(-64)=144+256=400\n\nt_1=(-b-\sqrt\Delta)/(2a);\ t_2=(-b+\sqrt\Delta)/(2a)\n\n\sqrt\Delta=√(400)=20\n\nt_1=(12-20)/(2\cdot1)=(-8)/(2)=-4 < 0;\ t_2=(12+20)/(2\cdot1)=(32)/(2)=16\n\ny^2=16\Rightarrow y=\pm√(16)\Rightarrow y=-4\ or\ y=4\n\nx=(8)/(-4)=-2\ or\ x=(8)/(4)=2\n\nSolution:x=-2\ and\ y=-4\ or\ x=2\ and\ y=4$

Given the graph what is a,h,k