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Lubov Fominskaja [6]
3 years ago
15

one container holds a 1 7/8 quarts of water and a second container holds 5 3/4 quarts of water. how many more quarts of water do

es the second container hold than the first container?
Mathematics
1 answer:
dimulka [17.4K]3 years ago
8 0
So lets solve this!

Quarts of water 1st container holds= 17/8

Quarts of water 2nd container holds= 5 3/4

Quarts of water does the second container holds than the first container=?

First of all we need to change 5 3/4 into an improper fraction

5 3/4= 23/4

Secondly, we need to find the LCM of 17/8 and 23/4

LCM:

17/8= 51/24

23/4= 138/24

So now we have got both of the LCM's 

Quarts of second container holds the first container= 23/4 - 138/24=87/24

Your answer is 87/24

Hope this Helps!




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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
If y = x - 6 were changed to y = x + 8, how would the graph of the new function compare with the first one?
11111nata11111 [884]
The first one would have a y-intercept of (0,-6) and the second one would have a y-intercept of (0,8) <span>and x (the slope) would still be the same.
</span><span>y=x-6 would have points with (0,-6) (2,-4) (4,-2) (6,0)
y=x+8 would have points with (0,8) (-2,6) (-4,4) (-6,2)
</span><span>In other words shifted up. Hope this helps!</span>

8 0
4 years ago
Read 2 more answers
WHICH ONE PLZ ANSWER
ValentinkaMS [17]

Answer:

area of a 1st

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A farmer bought 16 sheep, with 100 dollar insurance on each of them. He then bought a $25 heater. If he paid $925, how much did
lubasha [3.4K]
$46
heater and insurance for each sheep subtracted from the total amount leaves $740. Then divided by the amount of sheep equals $46!
Hope I helped!
4 0
3 years ago
Alessandro wrote the quadratic equation –6 = x2 + 4x – 1 in standard form. What is the value of c in his new equation?
Black_prince [1.1K]

Answer:

5

Step-by-step explanation:

To write it in standard form, we set the equation equal to 0.  To do this, we add 6 to each side:

-6+6 = x² + 4x - 1 + 6

0 = x² + 4x + 5

The related function is

y = x² + 4x + 5

The value of c in this function is 5.

7 0
3 years ago
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