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padilas [110]
3 years ago
12

Which is greater 7/9 or 85/100. Please help

Mathematics
2 answers:
zloy xaker [14]3 years ago
6 0
\frac{85}{100}= \frac{85:5}{100:5}= \frac{17}{20}= \frac{17\cdot9}{20\cdot9}= \frac{153}{180} \\\\ \frac{7}{9}= \frac{7\cdot20}{9\cdot20}= \frac{140}{180} \\\\ \frac{85}{100}>\frac{7}{9}

Ans.\ greater\ is\  \frac{85}{100}
Alika [10]3 years ago
4 0
\frac{7}{9} \ \ or \ \ \frac{85}{100} \\ \\ \frac{7}{9} \ \ or \ \ \frac{17}{20} \\ \\\frac{7\cdot 20}{180} \ \ or \ \ \frac{17\cdot 9}{180}\\ \\\frac{140}{180} \ \ < \ \ \frac{153}{180} \\ \\ Answer : \ \frac{7}{9} \ \ < \ \ \frac{85}{100}


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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
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