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2 years ago

Which of the following is an equivalent form of the compound inequality −33 > −3x − 6 ≥ −6?

Mathematics

1 answer:

wolverine [178]2 years ago

6 0

ANSWER

$9\: < x \leqslant0$

EXPLANATION

The given compound inequality is

$- 33 \: > - 3x - 6 \geqslant - 6$

We need to simplify this inequality so that we can obtain x standing alone between the inequality signs.

We add 6 through out the inequality.

$- 33 + 6\: > - 3x - 6 + 6 \geqslant - 6 + 6$

This simplifies to:

$- 27\: > - 3x \geqslant 0$

We now divide through by -3 and reverse the inequality sign.

$\frac{- 27}{ - 3} \: < \frac{ - 3x}{ - 3} \leqslant \frac{0}{ - 3}$

We now simplify to get:

$9\: < x \leqslant 0$

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A 4/7 inch pipe is to be shortened to 3/8 inch. How much must be removed?

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11/56 inches or 0.196 inches must be removed.

The pipe measures 4/7 inches but needs to be reduced to 3/8 inches.

In order to find out the inches to be removed, you must subtract the length that the pipe should be from the length that it currently is.

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Given: The coordinates of triangle PQR are P(0, 0), Q(2a, 0), and R(2b, 2c).

masha68 [24]

Answer:

The line containing the midpoints of two sides of a triangle is parallel to the third side ⇒ proved down

Step-by-step explanation:

* Lets revise the rules of the midpoint and the slope to prove the

problem

- The slope of a line whose endpoints are (x1 , y1) and (x2 , y2) is

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- The mid-point of a line whose endpoints are (x1 , y1) and (x2 , y2) is

$(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

* Lets solve the problem

- PQR is a triangle of vertices P (0 , 0) , Q (2a , 0) , R (2b , 2c)

- Lets find the mid-poits of PQ called A

∵ Point P is (x1 , y1) and point Q is (x2 , y2)

∴ x1 = 0 , x2 = 2a and y1 = 0 , y2 = 0

∵ A is the mid-point of PQ

∴ $A=(\frac{0+2a}{2},\frac{0+0}{2})=(\frac{2a}{2},\frac{0}{2})=(a,0)$

- Lets find the mid-poits of PR which called B

∵ Point P is (x1 , y1) and point R is (x2 , y2)

∴ x1 = 0 , x2 = 2b and y1 = 0 , y2 = 2c

∵ B is the mid-point of PR

∴ $B=(\frac{0+2b}{2},\frac{0+2c}{2})=(\frac{2b}{2},\frac{2c}{2})=(b,c)$

- The parallel line have equal slopes, so lets find the slopes of AB and

QR to prove that they have same slopes then they are parallel

# Slope of AB

∵ Point A is (x1 , y1) and point B is (x2 , y2)

∵ Point A = (a , 0) and point B = (b , c)

∴ x1 = a , x2 = b and y1 = 0 and y2 = c

∴ The slope of AB is $m=\frac{c-0}{b-a}=\frac{c}{b-a}$

# Slope of QR

∵ Point Q is (x1 , y1) and point R is (x2 , y2)

∵ Point Q = (2a , 0) and point R = (2b , 2c)

∴ x1 = 2a , x2 = 2b and y1 = 0 and y2 = 2c

∴ The slope of AB is $m=\frac{2c-0}{2b-2a}=\frac{2c}{2(b-c)}=\frac{c}{b-a}$

∵ The slopes of AB and QR are equal

∴ AB // QR

∵ AB is the line containing the midpoints of PQ and PR of Δ PQR

∵ QR is the third side of the triangle

∴ The line containing the midpoints of two sides of a triangle is parallel

to the third side

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