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aivan3 [116]
3 years ago
14

Find the quotient -72/-12=

Mathematics
2 answers:
LUCKY_DIMON [66]3 years ago
8 0

Answer: +6

Step-by-step explanation: To divide -72 by -12, it's important to understand that when we divide two negative numbers, we will always get a positive quotient.

So -72 ÷ -12 is +6.

natka813 [3]3 years ago
7 0

answer is -72/-12 = 6

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HELP ASAP (Geometry)
Andrei [34K]

1) Parallel line: y=-2x-3

2) Rectangle

3) Perpendicular line: y = 0.5x + 2.5

4) x-coordinate: 2.7

5) Distance: d=\sqrt{(4-3)^2+(7-1)^2}

6) 3/8

7) Perimeter: 12.4 units

8) Area: 8 square units

9) Two slopes of triangle ABC are opposite reciprocals

10) Perpendicular line: y-5=-4(x-(-1))

Step-by-step explanation:

1)

The equation of a line is in the form

y=mx+q

where m is the slope and q is the y-intercept.

Two lines are parallel to each other if they have same slope m.

The line given in this problem is

y=-2x+7

So its slope is m=-2. Therefore, the only line parallel to this one is the line which have the same slope, which is:

y=-2x-3

Since it also has m=-2

2)

We can verify that this is a rectangle by checking that the two diagonals are congruent. We have:

- First diagonal: d_1 = \sqrt{(-3-(-1))^2+(4-(-2))^2}=\sqrt{(-2)^2+(6)^2}=6.32

- Second diagonal: d_2 = \sqrt{(1-(-5))^2+(0-2)^2}=\sqrt{6^2+(-2)^2}=6.32

The diagonals are congruent, so this is a rectangle.

3)

Given points A (0,1) and B (-2,5), the slope of the line is:

m=\frac{5-1}{-2-0}=-2

The slope of a line perpendicular to AB is equal to the inverse reciprocal of the slope of AB, so:

m'=\frac{1}{2}

And using the slope-intercept for,

y-y_0 = m(x-x_0)

Using the point (x_0,y_0)=(7,1) we find:

y-1=\frac{1}{2}(x-7)

And re-arranging,

y-1 = \frac{1}{2}x-\frac{7}{2}\\y=\frac{1}{2}x-\frac{5}{2}\\y=0.5x-2.5

4)

The endpoints of the segment are X(1,2) and Y(6,7).

We have to divide the sgment into 1/3 and 2/3 parts from X to Y, so for the x-coordinate we get:

x' = x_0 + \frac{1}{3}(x_1 - x_0) = 1+\frac{1}{3}(6-1)=2.7

5)

The distance between two points A(x_A,y_A) and B(x_B,y_B) is given by

d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}

In this problem, the two points are

E(3,1)

F(4,7)

So the distance is given by

d=\sqrt{(4-3)^2+(7-1)^2}

6)

We have:

A(3,4)

B(11,3)

Point C divides the segment into two parts with 3:5 ratio.

The distance between the x-coordinates of A and B is 8 units: this means that the x-coordinate of C falls 3 units to the right of the x-coordinate of A and 5 units to the left of the x-coordinate of B, so overall, the x-coordinate of C falls at

\frac{3}{3+5}=\frac{3}{8}

of the  distance between A and B.

7)

To find the perimeter, we have to calculate the length of each side:

d_{EF}=\sqrt{(x_E-x_F)^2+(y_E-y_F)^2}=\sqrt{(-1-2)^2+(6-4)^2}=3.6

d_{FG}=\sqrt{(x_G-x_F)^2+(y_G-y_F)^2}=\sqrt{(-1-2)^2+(3-4)^2}=3.2

d_{GH}=\sqrt{(x_G-x_H)^2+(y_G-y_H)^2}=\sqrt{(-1-(-3))^2+(3-3)^2}=2

d_{EH}=\sqrt{(x_E-x_H)^2+(y_E-y_H)^2}=\sqrt{(-1-(-3))^2+(6-3)^2}=3.6

So the perimeter is

p = 3.6 + 3.2 + 2 + 3.6 = 12.4

8)

The area of a triangle is

A=\frac{1}{2}(base)(height)

For this triangle,

Base = XW

Height = YZ

We calculate the length of the base and of the height:

Base =XW=\sqrt{(x_X-x_W)^2+(y_X-y_W)^2}=\sqrt{(6-2)^2+(3-(-1))^2}=5.7

Height =YZ=\sqrt{(x_Y-x_Z)^2+(y_Y-y_Z)^2}=\sqrt{(7-5)^2+(0-2)^2}=2.8

So the area is

A=\frac{1}{2}(XW)(YZ)=\frac{1}{2}(5.7)(2.8)=8

9)

A triangle is a right triangle when there is one right angle. This means that two sides of the triangle are perpendicular to each other: however, two lines are perpendicular when their slopes are opposite reciprocals. Therefore, this means that the true statement is

"Two slopes of triangle ABC are opposite reciprocals"

10)

The initial line is

y=\frac{1}{4}x-6

A line perpendicular to this one must have a slope which is the opposite reciprocal, so

m'=-4

Using the slope-intercept form,

y-y_0 = m'(x-x_0)

And using the point

(x_0,y_0)=(-1,5)

we find:

y-5=-4(x-(-1))

Learn more about parallel and perpendicular lines:

brainly.com/question/3414323

brainly.com/question/3569195

#LearnwithBrainly

8 0
3 years ago
Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation o
Ierofanga [76]

Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, \mu = 5.5 mins

Standard deviation, \sigma = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒ z_1=\frac{x_1-\mu}{\sigma}   ...raw score,x_1 = 5.4

⇒ Plugging the values.

⇒ z_1=\frac{5.4-5.5}{0.4}

⇒ z_1=-0.25  

For raw score 5.5 the z score is.

⇒ z_2=\frac{5.8-5.5}{0.4}  

⇒ z_2=0.75

Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒ P(5.4

⇒ P(-0.25

⇒ P(z

⇒ z(1.5)=0.7734 and z(-0.25)=0.4013.<em>..from z -score table.</em>

⇒ 0.7734-0.4013

⇒ 0.3721

To find the percentage we have to multiply with 100.

⇒ 0.3721\times 100

⇒ 37.21 %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

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Answer: %0.00100010001

Step-by-step explanation:

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Answer:

140.80

Step-by-step explanation:

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