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3 years ago

Do the equations y=3x+5 and y=3x-5 have no, one or many solution(s)?

2 answers:

4 0

Y=3x+5
y=3x-5
(I remember that paralell lines have same slope and paralell lines with different y intercepts have no solutions so I wuld say no solutions)
I will solve
set them eaqual to each other
3x+5=y=3x-5
3x+5=3x-5
subtract 5 from both sides
3x=3x-10
subtract 3x from both sides
0=-10
false
there is no solution

Anarel [89]3 years ago

4 0

They will have no solutions. since the slope (3x) of both equations are the same and they have different y-intercepts (-5 and +5) these lines are parallel and will never intersect

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Yes, these 2 equations are equivalent

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3 years ago

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What is the slope of the line

cupoosta [38]

Answer:

I believe the slope would be -2

Step-by-step explanation:

The line goes <u>down 2</u> boxes and <u>over 1</u> box before it intersects with another point. Knowing slope = $\frac{rise}{run}$ , we know that slope = $\frac{-2}{1}$ which is equal to -2.

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3 years ago

Use the discriminant to describe the roots of each equation. Then select the best description. 2 = x2 + 5x. A) Double root B) Re

fredd [130]

$\bf 2=x^2+5x\implies 0=x^2+5x-2\\\\\\\qquad \qquad \qquad \textit{discriminant of a quadratic}\\\\\\0=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+5}x\stackrel{\stackrel{c}{\downarrow }}{-2}~~~~~~~~\stackrel{discriminant}{b^2-4ac}=\begin{cases}0&\textit{one solution}\\positive&\textit{two solutions}\\negative&\textit{no solution}\end{cases}\\\\\\(5)^2-4(1)(-2)\implies 25+8\implies 33$

so we have a 33, namely two real solutions for that quadratic.

usually that number goes into a √, if you have covered the quadratic formula, you'd see it there, namely that'd be equivalent to √(33), now 33 is a prime number, and √(33) is yields an irrational value, specifically because a prime number is indivisible other than by itself or 1.

so 33 can only afford us two real irrational roots.

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2 years ago

You have quarters and dimes that total $2.80. Your friend says it is possible that the number of quarters is 8 more than the num

AnnyKZ [126]

Answer:

Step-by-step explanation:

d = number of dimes

2.80 = 0.1d + 0.25 (d + 8 )

2.80 = 0.1d + 0.25d + 2 Combine like terms

2.80 = 0.35d + 2 Subtract 2 from both sides

0.80 = 0.35d Divide both sides by 0.35

d ≈ 2.29

No, it's not possible to have more quarters than dimes when the total is $2.80 because it's not possible to have part of a dime.

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2 years ago

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mel-nik [20]

Answer:6hrs and &270

8hours and &360

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Step-by-step explanation:

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2 years ago