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bonufazy [111]
2 years ago
7

After finishing the prep work, Gilberta and María start removing wallpaper at the same time. Gilberta removes the paper at a con

stant rate of
4.3

m
2
4.3m
2
4, point, 3, start text, m, end text, squared per hour, while María removes
3.4

m
2
3.4m
2
3, point, 4, start text, m, end text, squared of paper per hour. Gilberta's room starts with
35

m
2
35m
2
35, start text, m, end text, squared of paper, and María's room starts with
30.5

m
2
30.5m
2
30, point, 5, start text, m, end text, squared of paper.
Let
t
tt represent the time, in hours, since Gilberta and María start removing the wallpaper.
Complete the inequality to represent the times when Gilberta has more wallpaper left in her room than María has in hers.
t
tt

select inequality symbol


hours

Show Calculator
Mathematics
1 answer:
Alexxx [7]2 years ago
4 0

Using linear functions, the inequality that represents when Gilberta has more wallpaper left in her room than María has in hers is: t < 5.

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = mx + b

In which:

  • m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
  • b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

For this problem, we consider:

  • The initial amount as the y-intercept.
  • The rate as the slope.

Hence the amounts Gilberta and Maria have left after t hours are given by:

  • G(t) = 35 - 4.3t.
  • M(t) = 30.5 - 3.4t.

Gilberta has more papers when:

G(t) > M(t).

Hence:

35 - 4.3t > 30.5 - 3.4t

-0.9t < -4.5

Multiplying by -1:

0.9t < 4.5

t < 4.5/0.9

t < 5.

Hence the inequality is:

t < 5.

More can be learned about linear functions at brainly.com/question/24808124

#SPJ1

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Answer:

The area of rhombus PQRS is 120 m.

Step-by-step explanation:

Consider the rhombus PQRS.

All the sides of a rhombus are equal.

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The diagonals PR and QS bisect each other.

Let the point at of intersection of the two diagonals be denoted by <em>X</em>.

Consider the triangle QXR.

QR = 13 m

XR = 12 m

The triangle QXR is a right angled triangle.

Using the Pythagorean theorem compute the length of QX as follows:

QR² = XR² + QX²

QX² = QR² - XR²

       = 13² - 12²

       = 25

 QX = √25

       = 5 m

The measure of the two diagonals are:

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QS = 2 × QX = 2 × 5 = 10 m

The area of a rhombus is:

\text{Area}=\frac{1}{2}\times d_{1}\times d_{2}

Compute the area of rhombus PQRS as follows:

\text{Area}=\frac{1}{2}\times PR\times QS

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Thus, the area of rhombus PQRS is 120 m.

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3 years ago
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