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klemol [59]
3 years ago
15

A student concluded that the inequality −3+2y≤4x is equivalent to the inequality y≥2x+32, as shown below. Describe and correct t

he student’s error.
Mathematics
1 answer:
timofeeve [1]3 years ago
5 0

Answer:

-3 +2y \leq 4x

2y \leq 4x+3

y \leq 2x + \frac{3}{2}

And that's different from the claim of the student that:

y \geq 2x +\frac{3}{2}

The error of the student is that he/she changes the sign of the inequality from \leq to \geq and that's not possible since we don't multiply both sides of the equation by -1

Step-by-step explanation:

For this case we have the following inequality:

-3 +2y \leq 4x

We want to rewrite the last expression with y in the left and x in the right so we can begin adding 3 in both sides of the inequality and we got:

2y \leq 4x+3

Now we can divide both sides of the inequality by 2 and we got:

y \leq 2x + \frac{3}{2}

And that's different from the claim of the student that:

y \geq 2x +\frac{3}{2}

The error of the student is that he/she changes the sign of the inequality from \leq to \geq and that's not possible since we don't multiply both sides of the equation by -1

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3 years ago
Write the algebraic expression that is represented by the model
Salsk061 [2.6K]
3x+2y=5

Hop this helps
4 0
3 years ago
Read 2 more answers
Consider the region bounded by the curves y=|x^2+x-12|,x=-5,and x=5 and the x-axis
Tasya [4]
Ooh, fun

what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up

so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points

we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5


A.
\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx

B.
sepearte the integrals
\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}

next one
\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}

the last one you can do yourself, it is \frac{50}{3}
the sum is \frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}


so the area under the curve is \frac{233}{3}
6 0
3 years ago
Help!! (You will get 100 points! & Brainliest if you are correct)
Dafna1 [17]

Answer: The first number that appears in both sequences is 28.

Step-by-step explanation:

Let's write down numbers from each of the sequences

Sequence 1) We need to start from 7 and multiply 4

7x4=28, 28x4=112

The sequence is 7,28,112...

Sequence 2) We need to start from 8 and add 5

5+8=13, 13+5=18, 18+5=23, 23+5=28

The sequence is 8,13,18,23,28...

They both have 28

5 0
2 years ago
Read 2 more answers
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