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viva [34]
3 years ago
9

Jeremiah is working on a model bridge. He needs to create triangular components, and he

Mathematics
2 answers:
Darina [25.2K]3 years ago
6 0

Answer:

The answer is No, according to the Triangle Theorem.

Step-by-step explanation:

When you have the sides of the triangle, you must be able to add up 2 sides and have a larger number than the other side.

4+5 = 9 which is greater than 1

5+1 = 6 which is greater than 4

4+1=5 which is equal but not greater than 5

tekilochka [14]3 years ago
3 0

Answer:

No, he would not be able to create a triangle without  modifying the lengths of the toothpick

Step-by-step explanation:

In order to determine if the toothpicks can be used to create the triangle, make use of the Pythagoras theorem

The Pythagoras theorem : a² + b² = c²

where a = length

b = base

c =  hypotenuse

the square of the longest side of the triangle should be equal to the sum of the square of the two shortest sides

4² + 1²

16 + 1 = 15

5² = 25

15 is not equal to 25

So the lengths cannot be used

the 1 in length would need to be extended by 2 inches to be used

To confirm

3² + 5²

16 + 9 = 25

This is correct

In conclusion, in order to determine if the lengths can be used, check with Pythagoras theorem. If the lengths can be used,  the square of the longest side of the triangle should be equal to the sum of the square of the two shortest sides

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Alex787 [66]

A quadrilateral is any figure with 4 sides, no matter what the lengths of
the sides or the sizes of the angles are ... just as long as it has four straight
sides that meet and close it up.

Once you start imposing some special requirements on the lengths of
the sides,  or their relationship to each other, or the size of the angles,
you start making special kinds of quadrilaterals, that have special names.

The simplest requirement of all is that there must be one pair of sides that
are parallel to each other.  That makes a quadrilateral called a 'trapezoid'.
That's why a quadrilateral is not always a trapezoid.

Here are some other, more strict requirements, that make other special
quadrilaterals:

-- Two pairs of parallel sides . . . . 'parallelogram'

-- Two pairs of parallel sides
AND all angles the same size . . . . 'rectangle'
           (also a special kind of parallelogram)

-- Two pairs of parallel sides
AND all sides the same length . . . 'rhombus'
           (also a special kind of parallelogram)

-- Two pairs of parallel sides
AND all sides the same length
AND all angles the same size . . . . 'square'.
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5 0
3 years ago
(07.01 MC)
OverLord2011 [107]

Answer: 12/5

Step-by-step explanation:

4 0
3 years ago
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CAN SOMEONE PLEASE ANSWER THIS SOON! THANK YOU!
lyudmila [28]

We want to find one-half of the reciprocal of 7/sqrt(98). Let's write down an expression for this:

\dfrac{1}{2} \times \dfrac{\sqrt{98}}{7}

We can rewrite 98 into 2 \times 49

\dfrac{1}{2} \times \dfrac{\sqrt{2 \times 49}}{7}

=\dfrac{1}{2} \times \dfrac{\sqrt{2} \times \sqrt{49}}{7}

The square root of 49 is 7

=\dfrac{1}{2} \times \dfrac{\sqrt{2} \times 7}{7}

=\dfrac{1}{2} \times\sqrt{2}

=\dfrac{\sqrt{2}}{2}

This should be your answer. Let me know if you need any clarifications, thanks!

5 0
3 years ago
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How to create an equation with infinitely many solutions.
AlekseyPX

Answer:

4 x + 5 = 2 x + 2 x + 5

Step-by-step explanation:

If we simplify the following equation it will be 4 x + 5 = 4 x + 5. This equation has infinitely many solutions because every value for x we substitute in, it will be the same on both sides, for example

Substitute x = 7 into 4 x + 5 = 4 x + 5

4 × ( 7 ) + 5 = 4 × ( 7 ) + 5

28 + 5 = 28 + 5

33 = 33

Every x value that we substitute in will result in the equation having the same result on both sides

6 0
3 years ago
Can anyone please make up some easy questions for me to do? I need at least five
KatRina [158]

Answer:

well ig thats good

Step-by-step explanation:

thanks

5 0
3 years ago
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