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Vsevolod [243]
3 years ago
12

What is 120/160 in simplest form?

Mathematics
2 answers:
Oxana [17]3 years ago
8 0

Answer:

3/4

Step-by-step explanation:

divide both 120 nd 160 by 40

Ksju [112]3 years ago
4 0

Answer:

3/4

Step-by-step explanation:

120/160

12/16

3/4

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#14 pls! Help me please y’all
dsp73

Answer: The Answer is 7, Hope this helps! :)

Step-by-step explanation: We know that 6 and 12 are related, 6 times 2 is 12 so if you do the same for 4 and you multiply 4 by 2 you get 8 for the blank side, then subtract 1 from 8 to get 7. Same for the bigger triangle, 6 times 2 is 12 subtracted by 1 is 11.

So final answer is 7 for X!

4 0
3 years ago
Write an equation of the line with the given slope and y-intercept,(4,3) and (0,1)
PSYCHO15rus [73]

Answer:

y=1/2x+1

Step-by-step explanation:

First use slope formula.

\frac{y2-y1}{x2-x1}

Plug in the information needed.

\frac{1-3}{0-4}=\frac{-2}{-4}=\frac{1}{2}

The slope is \frac{1}{2}.

Now, use point-slope formula.

y-y1=m(x-x1)

Plug in the information needed.

y-3=1/2(x-4)

y-3=1/2x-2

y=1/2x+1

The equation of the line in slope-intercept form is y=1/2x+1.

Hope this helps!

If not, I am sorry.

5 0
2 years ago
Plz help ma :/<br><br><br> Helpppppppppppppppppppppp
Marizza181 [45]

Answer:

It is 432

Step-by-step explanation:

6 0
3 years ago
Questions 16-17 | Math 1 - 0 points Solve the graph Help needed !!
iris [78.8K]

Answer:

16) The area of the circle is 25.1 units²

17) JKLM is a parallelogram but not a rectangle

Step-by-step explanation:

16) Lets talk about the area of the circle

- To find the area of the circle you must find the length of the radius

- In the problem you have the center of the circle and a point on

 the circle, so you can find the length of the radius by using the

 distance rule

* Lets solve the problem

∵ The center of the circle is (1 , 3)

∵ The point on the circle is (3 , 5)

- Using the rule of the distance between two points

* Lets revise it

- The distance between the two points (x1 , y1) and (x2 , y2) is:

 Distance = √[(x2 - x1)² + (y2 - y1)²]

∴ r = √[(3 - 1)² + (5 - 3)²] = √[4 + 4] = √8 = 2√2 units

∵ The area of the circle = πr²

∴ The area of the circle = π (2√2)² = 8π = 25.1 units²

* The area of the circle is 25.1 units²

17) To prove a quadrilateral is a parallelogram, prove that every

     to sides are parallel or equal or the two diagonal bisect

     each other

* The parallelogram can be rectangle if two adjacent sides are

  perpendicular to each other (measure of angle between them is 90°)

 or its diagonals are equal in length

- The parallel lines have equal slopes, then to prove the

  quadrilateral is a parallelogram, we will find the slopes of

  each opposite sides

* Lets find from the graph the vertices of the quadrilateral

∵ J = (0 ,2) , K (2 , 5) , L (5 , 0) , M (3 , -3)

- The opposite sides are JK , ML and JM , KL

- The slope of any line passing through point (x1 , y1) and (x2 , y2) is

 m = (y2 - y1)/(x2 - x1)

∵ The slop of JK = (5 - 2)/(2 - 0) = 3/2 ⇒ (1)

∵ The slope of LM = (-3 - 0)/(3 - 5)= -3/-2 = 3/2 ⇒ (2)

- From (1) and (2)

∴ JK // LM

∵ The slope of KL = (0 - 5)/(5 - 2) = -5/3 ⇒ (3)

∵ The slope of JM = (-3 - 2)/(3 - 0)= -5/3 ⇒ (4)

- From (3) and (4)

∴ KL // JM

∵ Each two opposite sides are parallel in the quadrilateral JKLM

∴ It is a parallelogram

- The product of the slopes of the perpendicular line is -1

* lets check the slopes of two adjacent sides in the JKLM

∵ The slope of JK = 3/2 and the slope of KL = -5/3

∵ 3/2 × -5/3 = -5/2 ≠ -1

∴ JKLM is a parallelogram but not a rectangle

4 0
3 years ago
Two objects were lifted by a machine. One object had a mass of 2 kilograms and was lifted at a speed of 2m/sec. the other had a
OLga [1]
Sadly, after giving all the necessary data, you forgot to ask the question.
Here are some general considerations that jump out when we play with
that data:

<em>For the first object:</em>
The object's weight is (mass) x (gravity) = 2 x 9.8 = 19.6 newtons
The force needed to lift it at a steady speed is 19.6 newtons.
The potential energy it gains every time it rises 1 meter is 19.6 joules.
If it's rising at 2 meters per second, then it's gaining 39.2 joules of
     potential energy per second.
The machine that's lifting it is providing 39.2 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(2)(4) = 4 joules.

<em>For the second object:</em>
The object's weight is (mass) x (gravity) = 4 x 9.8 = 39.2 newtons
The force needed to lift it at a steady speed is 39.2 newtons.
The potential energy it gains every time it rises 1 meter is 39.2 joules.
If it's rising at 3 meters per second, then it's gaining 117.6 joules of
     potential energy per second.
The machine that's lifting it is providing 117.6 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(4)(9) = 18 joules.

If you go back and find out what the question is, there's a good chance that
you might find the answer here, or something that can lead you to it.

4 0
3 years ago
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