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never [62]
1 year ago
11

mrs jones gave 1/5 of a cake to her neighbour. she then sliced the remainder into 3 pieces for her 3 children, ken, ron and davi

d, in the ratio 1 : 3 : 4. what is the fraction of the cake was ron's pieces
Mathematics
1 answer:
Arturiano [62]1 year ago
4 0

The fraction of the cake that was ron's pieces is 1/5.

<h3>What is the fraction of the cake was ron's pieces?</h3>

Ratio demonstrates how many times one number can fit into another number. Ratios contrast two numbers by ordinarily dividing them. A/B will be the formula if one is comparing one data point (A) to another data point (B).

Since Mrs Jones gave 1/5 of a cake to her neighbour, the remaining fraction will be:

= 1 - 1/5

= 4/5

The fraction of the cake that was ron's pieces will be:

= 2/(1+3+4) × 4/5

= 2/8 × 4/5

= 8/40

= 1/5

Learn more about ratio on:

brainly.com/question/2328454

#SPJ1

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A ticket to a movie theatre for a student is $7. The cost for an adult is $2 more than for a student. How much would it cost 13
MakcuM [25]

Answer: it would cost a total of $194

Step-by-step explanation:

A ticket to a movie theatre for a student is $7. The cost for an adult is $2 more than for a student.

If student ticket costs $x, adult ticket would cost x + 2

This means that the cost for an adult is

7 + 2 = $9

The cost of 13 adult tickets to the movie theatre would be

13 × 9 = $117.

The cost of 11 student tickets to the movie theatre would be

11 × 7 = $77

Total cost of 13 adult tickets and 11 student tickets would be

117 + 77 = $194

3 0
3 years ago
The measure of an angle in standard position is given. Find two positive angles and two negative angles that are coterminal with
Pani-rosa [81]
Notice the picture below

negative angles, are just angles that go "clockwise", namely, the same direction a clock hands move hmmm so....  and one revolution is just 2π

now, you can have angles bigger than 2π of course, by simply keep going around, so, if you go around 3 times on the circle, say "counter-clockwise", or from right-to-left, counter as a clock goes, 3 times or 3 revolutions will give you an angle of 6π, because 2π+2π+2π is 6π

now... say... you have this angle here... let us find another that lands on that same spot

by simply just add 2π to it :)  

\bf \cfrac{7}{6}+2=\cfrac{19}{6}\qquad thus\qquad \cfrac{7\pi} {6}+2\pi =\cfrac{19\pi }{6}&#10;\\\\\\&#10;\cfrac{19\pi }{6}\impliedby co-terminal\ angle

now, that's a positive one
and  \bf \cfrac{7}{6}-2=-\cfrac{5}{6}\qquad thus\qquad \cfrac{7\pi} {6}-2\pi =-\cfrac{5\pi }{6}&#10;\\\\\\&#10;-\cfrac{5\pi }{6}\impliedby \textit{is also a co-terminal angle}

to get more, just keep on subtracting or adding 2π


8 0
3 years ago
(2*)2-3×2*+2=0<br>4m-15°(×)m+75°​
Valentin [98]

Answer:

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

7 0
2 years ago
Is it C ? am I right
sveta [45]

Answer:

You are right!!

Step-by-step explanation:

- 8 = 4 x (- 2)

- 16 = 4 x (- 4)

- 24 = 4 x (- 6)

8 0
3 years ago
Solve the following inequalities 7 x minus 5 / 8 x + 3 &gt;4<br>​
olchik [2.2K]

Answer:

x> \frac{8}{51}

Step-by-step explanation:

7x -  \frac{5}{8} x + 3>4

Bring constants to one side, simplify:

\frac{51}{8} x>4 - 3 \\  \frac{51}{8} x>1 \\ x>1 \div  \frac{51}{8}  \\ x>1 \times  \frac{8}{51}  \\ x> \frac{8}{51}

*Note that the inequality sign only changes when you divide the whole inequality by a negative number.

7 0
3 years ago
Read 2 more answers
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