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Kazeer [188]
3 years ago
8

Five less than a number is greater than 21five less than a number is greater than 21​

Mathematics
1 answer:
Sophie [7]3 years ago
3 0
Sorry can u be more specific thank you
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How many different triangles can be constructed with three sides, each measuring 5 centimeters?
Akimi4 [234]
The first thing we will do is define an equilateral triangle:
 In geometry, an equilateral triangle is a regular polygon with three equal sides. In traditional Euclidean geometry, equilateral triangles are also equiangular, that is, the three internal angles are also congruent to each other, each angle with a value of 60 °
 Every equilateral triangle consists of three equal sides and three congruent angles.
 Therefore, there can be a triangle with three equal sides (5 centimeters in this case).
 Answer:
 1) one
5 0
3 years ago
How do i solve this question 58−(14)2
Naily [24]

Answer:

PEMDAS RULE

Step-by-step explanation:

STEP one is multiplication 58-14x2=58-28

STEP two is subtraction 58-28=30

Parenthesis

Exponent

Multiplication

Division

Addition

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3 0
3 years ago
Read 2 more answers
What is the y-intercept of the function y= -1/2cos(3/2x)
Flauer [41]

Answer: (0,-1/2)

Step-by-step explanation:

Edge :-)

5 0
3 years ago
What is a time in the past that you celebrated something good that happened? its for my english class, and i dont know what do w
lapo4ka [179]

Answer:

Easter, Christmas, birthdays, and New Year

Step-by-step explanation:

For me, I would choose Easter because before easter we always go to a church and stay there whole night. After 3-4 a.m, we go home and take a nap. But the most fun activity that we do during Easter is playing a game where we pick the best eggs and beat with each other. Later, our parents hide eggs and a couple of toys in the park and me and my siblings begin finding them.

3 0
3 years ago
IM AM SO CONFUSED
Semenov [28]

Hello,

Answer (1/2,-1/4)

8x^2-2x-1=0\\\\8x^2-4x+2x-1=0\\\\4x(2x-1)+2x-1=0\\\\(2x-1)(4x+1)=0\\\\sol=\{\dfrac{1}{2} ,-\dfrac{1}{4} \}\\

6 0
3 years ago
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