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Digiron [165]
3 years ago
8

During the first week of school Andrew purchased two pounds of Jellies and two pounds of Bonbons for $38. The second week of sch

ool he purchased two pounds of Jellies and four pounds of Chocolates for $74 and the third week he purchased two pounds of Jellies, two pounds of Bonbons and one pound of Chocolates for $53. What is the cost of one pound of Chocolates?
Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
7 0

Answer:

j = $7

b = $12

c = $15

Step-by-step explanation:

Let

j = cost of one pound of jelly

b = cost of one pound of Bonbon

c = cost of one pound of chocolate

First week:

two pounds of Jellies and two pounds of Bonbons for $38.

2j + 2b = 38

Second week:

two pounds of Jellies and four pounds of Chocolates for $74

2j + 4c = 74

Third week:

two pounds of Jellies, two pounds of Bonbons and one pound of Chocolates for $53.

2j + 2b + c = 53

Recall,

2j + 2b = 38

So,

38 + c = 53

c = 53 - 38

= 15

c = $15

Substitute c = 15 into

2j + 4c = 74

2j + 4(15) = 74

2j + 60 = 74

2j = 74 - 60

2j = 14

j = 14/2

= 7

j = $7

Substitute j = 7 into

2j + 2b = 38

2(7) + 2b = 38

14 + 2b = 38

2b = 38 - 14

2b = 24

b = 24/2

= 12

b = $12

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Answer:

<h2>b=-4</h2>

Step-by-step explanation:

<h3>-1/2b+3=5</h3><h3>-1/2=5-3</h3><h3> -1/2b=2×-2</h3><h2> b=-4</h2>
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3 years ago
What is the prime factorization of 891
Advocard [28]

891

prime factors   number to factorize

3                          297

3                          99

3                          33

3                          11

11                          1

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3 years ago
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creativ13 [48]
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3 years ago
Consider the line which passes through the point P(−1,-3,5), and which is parallel to the line x=1+7t, y=2+2t, z=3+4t Find the p
bija089 [108]

The given line is parameterized by

x(t) = 1 + 7t

y(t) = 2 + 2t

z(t) = 3 + 4t

and points in the same direction as the vector

d/dt (x(t), y(t), z(t)) = (7, 2, 4)

So, the line we want has parameteric equations

x(t) = -1 + 7t

y(t) = -3 + 2t

z(t) = 5 + 4t

Solve for t when one of x, y, or z is equal to 0 - this will tell you for which value of t the line cross a given plane. Then determine the other coordinates of these intersections.

• x = 0, which corresponds to the y-z plane:

0 = -1 + 7t   ⇒   7t = 1   ⇒   t = 1/7

y(1/7) = -3 + 2/7 = -19/7

z(1/7) = 5 + 4/7 = 39/7

⇒   intersection = (0, -19/7, 39/7)

• y = 0 (x-z plane):

0 = -3 + 2t   ⇒   2t = 3   ⇒   t = 3/2

x(3/2) = -1 + 21/2 = 19/2

z(3/2) = 5 + 12/2 = 11

⇒   intersection = (19/2, 0, 11)

• z = 0 (x-y plane):

0 = 5 + 4t   ⇒   4t = -5   ⇒   t = -5/4

x(-5/4) = -1 - 35/4 = -39/4

y(-5/4) = -3 - 10/4 = -11/2

⇒   intersection = (-39/4, -11/2, 0)

8 0
2 years ago
The figure below shows a quadrilateral ABCD. Sides AB and DC are congruent and parallel: A quadrilateral ABCD is shown with the
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A Quadrilateral A B C D in which Sides AB and DC are congruent and parallel.

The student has written the following explanation

Side AB is parallel to side DC so the alternate interior angles, angle ABD and angle BDC, are congruent. Side AB is equal to side DC and DB is the side common to triangles ABD and BCD. Therefore, the triangles ABD and CDB are congruent by SAS.

The student has also written

angles DBC and ADB are congruent and sides AD and BC are congruent. Angle DBC and angle ADB form a pair of alternate interior angles. Therefore, AD is congruent and parallel to BC. Quadrilateral ABCD is a parallelogram because its opposite sides are equal and parallel.

Postulate SAS completely describes the student's proof.

Because if in a quadrilateral one pair of opposite sides are equal and parallel then it is a parallelogram.

 

6 0
3 years ago
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