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2 years ago

Please answer correctly !!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Nikolay [14]2 years ago

4 0

Answer:

the last one [d)]

Step-by-step explanation:

First, we need to use the triangle inequality theorem. The sum of the lengths of the two sides must be greater than the third side.

a) 7+14=21=21 (this cannot represent a triangle)

b) 13+23=36<37 (this cannot represent a triangle)

c) 7+18=25=25 (this cannot represent a triangle)

d) 11+24=35>32 (this can represent a triangle)

d) is the answer

Hope this helps!

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10)

Oxana [17]

Answer:

I think it is 2 hours

Step-by-step explanation:

because if you have a whole 3 more hours of work you will get 900 bricks layd and if you minus a hour it might be free srry if it is wrong im kinda dumb

8 0

3 years ago

Five times the sum of a number and 27 is greater than or equal to six times the sum of that num

stepladder [879]

Question:

Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26.

What is the solution set of this problem?

Answer:

$x \leq -21$

Step-by-step explanation:

Given

<em>Represent the number with x</em>

So:

$5 * (x + 27) \geq 6 * (x + 26)$

Required

Determine the solution set

$5 * (x + 27) \geq 6 * (x + 26)$

Open Both Brackets

$5 * x + 5 * 27 \geq 6 *x + 6 * 26$

$5x + 135 \geq 6x + 156$

Collect Like Terms

$5x - 6x \geq 156-135$

$- x \geq 21$

Multiply both sides by -1

$x \leq -21$

Hence, the solution set is $x \leq -21$

7 0

3 years ago

The larger of two numbers is five more than twice the smaller number. The sum of the two numbers is 38. Find the numbers.

White raven [17]

Answer:

11 and 16

Step-by-step explanation:

A good way to solve these is to work backwards.

X= larger and Y= smaller

if X is 5 more than 2Y than subtract 5 from the total

38 - 5 = 33

divide by 3

33/3 = 11

11 = Y and double that is 22 so add the five back into X and you get 16

8 0

3 years ago

Which anwser choice would it be?

barxatty [35]

A is the answer for this one

4 0

3 years ago

Three consecutive integers have a sum of 45

monitta

Three consecutive integers that have a sum of 45 are 14, 15 and 16. 14+15+16=45

4 0

3 years ago

Read 2 more answers