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erik [133]
3 years ago
15

Denise bought a box of 560 pieces of candy.4/5 of the pieces of the candy were fruity and 1/4 of the remaining pieces of candy w

ere mint.How many of candy of the other flavors were there?
Mathematics
1 answer:
Liula [17]3 years ago
8 0

Answer:

<em>Number of Other flavor candy is 84.</em>

Step-by-step explanation:

Given that

A total of 560 pieces of candy are there in the box.

\frac{4}{5} of the pieces are fruity.

Number of Fruity pieces:

\dfrac{4}{5} \times 560\\\Rightarrow 4 \times 112 =448

Now, remaining pieces = Total pieces - Fruity pieces

remaining pieces = 560 - 448 = 112

Also given that \frac{1}4 of the remaining pieces are mind.

Number of mint pieces:

\dfrac{1}{4}\times 112 = 28

Other flavors remaining = Total pieces - Fruity pieces - Mint pieces

Other flavors remaining = 560 - 448 - 28

Other flavors remaining = 112 - 28 = <em>84</em>

<em>Number of Other flavor candy is 84.</em>

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Answer:

70%

Step-by-step explanation:

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After how many years are the salaries offered by company a and company b the same
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24000 + 3000x = 30000 + 2400x
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6 0
2 years ago
the class had a pizza party 3/8 of a pizza was left over and 4/8 of a pizza with other pizza was left over you put them both int
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You would have 7/8 of pizza slices total.
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3 years ago
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2 Here are two equations:
MatroZZZ [7]

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Step-by-step explanation:

a. Decide whether each neither of the equations,

i (3,4)

ii. (4,2.5)

ill. (5,5)

iv. (3,2)

To decide whether each point is solution to equations or not we will put the point in the equations

Equations are:

Equation 1: 6x + 4y = 34

Equation 2: 5x – 2y = 15

<u>i (3,4) </u>

Putting in Equation 1:

6(3) + 4(4) = 34\\18+16=34\\34=34\\

Putting in Equation 2:

5(3) - 2(4) = 15\\15-8 = 15\\7\neq 15

<u>ii. (4,2.5)</u>

Putting in Equation 1:

6(4) + 4(2.5) = 34\\24+10=34\\34=34\\

Putting in Equation 2:

5(4) - 2(2.5) = 15\\20-5 = 15\\15=15

<u>ill. (5,5)</u>

6(5) + 4(5) = 34\\30+20=34\\50\neq 34

Putting in Equation 2:

5(5) - 2(5) = 15\\25-10 = 15\\15=15

<u>iv. (3,2)</u>

6(3) + 4(2) = 34\\18+8=34\\26\neq 34

Putting in Equation 2:

5(3) - 2(2) = 15\\15-4 = 15\\11\neq 15

Hence,

(3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. Is it possible to have more than one (x, y) pair that is a solution to both

equations?

The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution

So,

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Keywords: Linear equations, Ordered pairs

Learn more about linear equations at:

  • brainly.com/question/10534381
  • brainly.com/question/10538663

#LearnwithBrainly

8 0
3 years ago
One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times sm
sergey [27]

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel

<h3><u>Solution:</u></h3>

Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.  

We have to prove that the lines are parallel.

If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.

Now, the 1st angle will be 1/6 of right angle is given as:

\begin{array}{l}{\rightarrow 1^{\text {st }} \text { angle }=\frac{1}{6} \times 90} \\\\ {\rightarrow 1^{\text {st }} \text { angle }=15 \text { degrees }}\end{array}

And now, 15 degrees is 11 times smaller than the other  

Then other angle = 11 times of 15 degrees

\text {Other angle }=11 \times 15=165 \text { degrees }

Now, sum of angles = 15 + 165 = 180 degrees.

As we expected their sum is 180 degrees. So the lines are parallel.

Hence, the given lines are parallel

5 0
3 years ago
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