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Temka [501]
3 years ago
11

I will mark you brainlest if you get this question correct

Mathematics
1 answer:
Luda [366]3 years ago
7 0

Answer:

12, 17, 30.

Step-by-step explanation:

Given

(a) to (d)

Required

Which cannot represent a triangle

The rule to use is:

a + b > c

So, we have: 12, 17, 30.

12 + 17 > 30 ===> 29 > 30 --- False

12 + 30 > 17 ==> 43 > 17

17 + 30 > 12

Because, one of the inequalities is False, then this can not represent the sides of a triangle.

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Sort each equation according to whether it has one solution, infinitely many solutions, or no solution.
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《ANSWER》

《LINEAR EQUATIONS》

Solving all we get as

↪1) 5 (x -2) = 5x - 7

since after solving X is cancelled , NO SOLUTION

↪2) -3 (x - 4) = -3x + 12

SINCE after solving X is any value , infinitely many solutions

↪3) 4 (x + 1) = 3x + 4

solving we get as , 4x+ 4 = 3x + 4 =》 X =0

only one solution

↪4) -2 (x-3) = 2x - 6

Only one solution

↪5) 6 (x + 5) = 6x + 11

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All the equations are solved by the distributive law of algebra
5 0
3 years ago
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Viktor [21]
The answer is B because math
5 0
3 years ago
Plz help I’m getting timed
Fed [463]

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6 0
3 years ago
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6h + 4 = 8 solve for h
Greeley [361]

Answer: h=4/6

Step-by-step explanation:

6h=8-4

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6 0
3 years ago
Hamburgers cost $2.50 and cheeseburgers cost $3.50 at a snack bar. Ben has sold no more than $30 worth of hamburgers and cheeseb
Irina18 [472]

Given 2.50x + 3.50y < 30.

Where x represent the number of hamburgers and y represent the number of cheeseburgers.

Now question is to find the maximum value of hamburgers Ben could have sold when he has sold 4 cheeseburgers.

So, first step is to plug in y=4 in the given inequality. So,

2.50x+3.50(4)<30

2.50x+14 <30

2.50x<30- 14 Subtracting 14 from each sides.

2.50x< 16

\frac{2.50x}{2.50} Dividing each sides by 2.50.

x<6.4

Now x being number of hamburgers must be an integer , so tha maximum value of x can be 6,

thus x = 6 hamburgers

So, the maximum value of hamburgers Ben could have sold is 6*2.5=$15

Hope this helps!!

8 0
3 years ago
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