What is 6 times 2/3 (two thirds)?

Answers

Answer 1
Answer: 4. 6 times 2 is 12 and 12 divided by 3 is 4
Answer 2
Answer:

Final answer:

The answer to the mathematical problem 6 times 2/3 is 4. This is solved by turning the whole number 6 into a fraction (6/1), multiplying the numerators together (6 times 2 equals 12), multiplying the denominators together (1 times 3 equals 3), and putting the new numerator over the new denominator (12/3), which simplifies to 4.

Explanation:

The question is asking you to multiply 6 by 2/3. Let's follow the steps to solve this multiplication problem.

  1. Turn the whole number, in this case 6, into a fraction by putting it over 1. This gives you 6/1.
  2. Multiply the top numbers (the numerators) together to get the new numerator. 6 * 2 = 12.
  3. Multiply the bottom numbers (the denominators) together to get the new denominator. 1 * 3 = 3.
  4. Put the new numerator over the new denominator to get your answer. 12/3 = 4.

So, 6 times 2/3 equals 4.

Learn more about Multiplication here:

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M(5, 7) is the midpoint of mc140-1.jpg. The coordinates of S are (6, 9). What are the coordinates of R? (4, 5) (7, 11) (5.5, 8) (10, 14)

Answers

Answer:

(4,5)

Step-by-step explanation:

Given that M is the midpoint of the Line RS. Where the coordinates of M, R and S are

M(5,7)

R(X,Y)

S(6,9)

Here we have to find the valuer of X and Y

We will use the mid point formula which is given below.

x=(x_1+x_2)/(2) \ny=(y_1+y_2)/(2)\n

Let put the coordinated of M and S and find coordinates of R

x=(x_1+x_2)/(2) \n5=(X+6)/(2) \n10=X+6\nX=4

y=(y_1+y_2)/(2)\n7=(Y+9)/(2) \n14=Y+9\ny=5

Hence our coordinates are

(4,5)

the coordinate of R(4,5)

In parallelogram ABCD, E is the midpoint of AB and F is the midpoint of DC . Let G be the intersection of the diagonal DB and the line segment EF . Prove that G is the midpoint of EF.

Answers

The midpoint of the line \overline{EF} is the point that divides \overline{EF} in two halves of the same length.

  • ΔDFG ≅ ΔBGE and \overline{FG}\overline{EG} by CPCTC, therefore, Gis the midpoint of \overline{EF}

Reasons:

The given parameters are;

The midpoint of AB in parallelogram ABCD = E

The midpoint of DC = F

Point of intersection of EF and DB = Point G

Required:

To prove that point G is the midpoint of EF.

Solution:

Statement         {}                       Reason

1. m∠BDC ≅ m∠ABD          {}  1. Alternate angles theorem

2. m∠DGF ≅ m∠BGE           {}2.Vertical angles theorem

3. \overline{DC} = \overline {AB}          {}                  3. Opposite sides of a parallelogram ABCD

4. \overline{CF}\overline{DF}          {}                 4. Definition of midpoint of DC

5. \overline{CF} = \mathbf{\overline{DF}}          {}                  5. Definition of congruency

6. \overline{CF} + \overline{DF} = DC         {}         6. Segment addition property

7. \overline{CF} + \overline{CF} = DC         {}          7. Substitution property

8. 2·\overline{CF} = DC        {}                 8. Addition

9. \overline{CF} = 0.5· \overline{DC} = \overline{DF}        {}  9. Division property    

Similarly;

10. \overline{AE} = 0.5·\overline{AB} = \overline{EB}         {}  10. Division property

11. 0.5· \overline{DC} = 0.5·\overline{AB}         {}     11. Multiplication property of equality

12. \overline{AE} = \overline{EB}          {}                 12. Substitution property

13. ΔDFG ≅ ΔBGE     {}             13. Angle-Angle-Side rule of congruency

14. \overline{FG}\overline{EG}                 {}          14. CPCTC   {}  

15. \overline{FG} = \overline{EG}     {}                       15. Definition of congruency

16. Point G is the midpoint of \overline{EF}{}  17. Definition of midpoint

Learn more about the midpoint of a line here:

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Answer:

GF = GE that prove G is the mid-point of EF

Step-by-step explanation:

In the Parallelogram ABCD

∵E is the mid-point of AB

∵F is the mid-point of CD

∵AB = CD opposite sides in the parallelogram

∴EB = DF⇒(1)

∵AB // CD opposite sides in the parallelogram

∴m∠EBD = m∠FDB alternate angles ⇒(2)

∵BD intersects EF at G

∴m∠BGE = m∠DGF vertically opposite angles ⇒(3)

By using (1) , (2) and (3) you can prove:

ΔBGE is congruent to ΔDGF ⇒ AAS

∴GF = GE

∴G is the mid-point of EF

if julie ran 3.6 miles from home to the gym and then walked 0.7 miles to the store she then ran from the store home if her total distance covered after she returned from the store was 7.2 miles how far is it from the store to her home? write and simplify and expression that would represent the distance she covered.

Answers

So,

Home ---- gym = 3.6 mi.
gym   ---- store = .7 mi.
store ----- home = x

3.6 + .7 + x = 7.2 mi.

Collect Like Terms

4.3 + x = 7.2

Subtract 4.3 from both sides
x = 2.9

The distance from the store to her home is 2.9 miles.

-9x+y=16
x-y=0

I have to solve by elimination.
Will someone help me please?

Answers

To do elimination, you combine the two equations by adding
  -9x + y = 16
+  x  -  y  = 0
 -8x  + 0y = 16

Now that there are 0y, we can solve for x:
-8x = 16
divide both sides by -8 
x = -2

To solve for y, plug the answer you got for x in to the original equations.

-9x + y = 16
-9(-2) + y = 16
18 + y = 16
y = -2 

x - y = 0
-2 - y = 0
y = -2

x = -2 and y = -2

-9x+y=16
x-y=0
add the two equations, the y's cancel each other out 
-8x=16
divide by -8 
x=-2

Assume that random guesses are made for nine multiple choice questions on an SAT​ test, so that there are nequals9 ​trials, each with probability of success​ (correct) given by p equals 0.35. Find the indicated probability for the number of correct answers. Find the probability that the number x of correct answers is fewer than 4.

Answers

Answer:  Find the probability that the number x of correct answers is fewer than 4 = 0.6087

Step-by-step explanation: Please find the attached files for the solution

I have a triangle with side lengths of 2 cm, 3 cm, and 4 cm. If I have a second triangle that is similar that has a shortest side length of 4 cm, what is the longest side length?

Answers

so 2,3,4
another one is
4,x,y
the shortest is 2 and 4
4/2=2

2 time 2=4

the other triangle sides area obviously twice as large

2 times 2=4
2 times 3=6
2 times 4=8



the longest sides is 8 cm