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Nataly [62]
3 years ago
5

Classify triangle ABC by its sides if side AB = 6x, side AC = 4x + 6 and side BC = 8x + 3. The perimeter of triangle ABC is 63.

Show all your work
Mathematics
2 answers:
Bad White [126]3 years ago
8 0

Answer:

AB = 18

AC = 18

BC = 27

Step-by-step explanation:

Solve for x first:

6x + 4x + 6 + 8x + 3 = 63

6x + 4x + 8x + 6 + 3 = 63

18x + 9 = 63

18x = 63 - 9

18x = 54

x = 54 / 18

x = 3

Now substitute x to all sides.

AB = 6x

6(3) = 18

AC = 4x + 6

4(3) + 6

12 + 6 = 18

BC = 8x + 3

8(3) + 3

24 + 3 = 27

18 + 18 + 27 = 63

63 = 63

sladkih [1.3K]3 years ago
3 0
6x + 4x+6 + 8x+3 = 63
6x+4x+8x = 63 - 3 - 6
18x = 54
x = 54/18
x = 3

Side AB = 6(3) = 18
Side AC = 4(3)+6 = 18
Side BC = 8(3)+3 = 27
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Questions (no partial grades if you don't show your work) 1. In a group of 6 boys and 4 girls, four children are to be selected.
mariarad [96]

Answer:

Total number of ways will be 209

Step-by-step explanation:

There are 6 boys and 4 girls in a group and 4 children are to be selected.

We have to find the number of ways that 4 children can be selected if at least one boy must be in the group of 4.

So the groups can be arranged as

(1 Boy + 3 girls), (2 Boy + 2 girls), (3 Boys + 1 girl), (4 boys)

Now we will find the combinations in which these arrangements can be done.

1 Boy and 3 girls = ^{6}C_{1}\times^{4}C_{3}=6\times4=24

2 Boy and 2 girls=^{6}C_{2}\times^{4}C_{2}=\frac{6!}{4!\times2!}\times\frac{4!}{2!\times2!}=15\times6=90

3 Boys and 1 girl = ^{6}C_{3}\times^{4}C_{1}=\frac{6!}{4!\times2!}\times\frac{4!}{3!}=\frac{6\times5\times4}{3 \times2} \times4=80

4 Boys = ^{6}C_{4}=\frac{6!}{4!\times2!} =\frac{6\times 5}{2\times1}=15

Now total number of ways = 24 + 90 + 80 + 15 = 209

7 0
3 years ago
What is the square root of 999,999,999,999,999,999,999,999,999,999,999
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Answer:

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Step-by-step explanation:

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Which of these sets could represent the side lengths of a right triangle?
Jlenok [28]

Answer:

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Step-by-step explanation:

In a right triangle:

(Base)^{2}  + (Perpendicular)^{2}   = (Hypotenuse)^{2}

Now, in the given triplets:

(a) {4, 8, 12}

Here, (4)^{2}  + (8)^{2}   = 16 + 64  = 80\\\implies H = \sqrt{80}  =  8.94

So, third side of the triangle   8.94 ≠ 12

Hence,  {4, 8, 12} is NOT a triplet.

(b) {6, 8, 10}

Here, (6)^{2}  + (8)^{2}   = 36 + 64  = 100\\\implies H = \sqrt{100}  =  10

So, third side of the triangle  10

Hence,  {6, 8, 10} is  a triplet.

(c) {6, 8, 15}

Here, (6)^{2}  + (8)^{2}   = 36 + 64  = 100\\\implies H = \sqrt{100}  =  10

So, third side of the triangle  10  ≠ 15

Hence,  {6, 8, 15} is  NOT a triplet.

(d) {5, 7, 13}

Here, (5)^{2}  + (7)^{2}   = 25 + 49  = 74\\\implies H = \sqrt{74}  =  8.60

So, third side of the triangle  8.60  ≠ 13

Hence, {5, 7, 13} is  NOT a triplet.

4 0
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