All categories

3 years ago

Will side lengths of 3 cm, 4 cm, and 5 cm form a triangle? explain.

Mathematics

2 answers:

Gelneren [198K]3 years ago

6 0

Yes, that is the famous first Pythagorean triplet: 3²+4²=5²

For three lengths to be able to form a triangle, it suffices that every one of those lengths is shorter than the sum and longer than the difference of the other two.

It is enough to check just one side.

So,

3 + 4 > 5

4 - 3 < 5

Your triangle is constructible.

just olya [345]3 years ago

4 0

Answer: yes because 3cm, 4cm, and 5cm form a triangle

Step-by-step explanation:

Comment the right answer if i'm wrong or not

You might be interested in

Helllllppppppppppp ....................

valina [46]

The correct answer is 3×10^16

5 0

3 years ago

Read 2 more answers

An empty suitcase has a mass of 2 kilograms. Draw a bar model to find its mass in grams.

miss Akunina [59]

Answer:

Since I can't draw it I'll give you the answer. 2,000 g.

Step-by-step explanation:

A kilogram is 1000 grams. Multiply 1000x2 to get 2000. But add the g for grams or the answer is considered wrong by teachers.

8 0

3 years ago

The following integral requires a preliminary step such as long division or a change of variables before using the method of par

shtirl [24]

Division yields

$\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}$

Now for partial fractions: you're looking for constants a, b, and c such that

$\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}$

$\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a$

which gives a + b = 2, c = 0, and 2a = -7, so that a = -7/2 and b = 11/2. Then

$\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}$

Now, in the integral we get

$\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx$

The first two terms are trivial to integrate. For the third, substitute y = x ² + 2 and dy = 2x dx to get

$\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}$

7 0

2 years ago

Help help help help help help help MATH MATH

kondaur [170]

Answer:

Step-by-step explanation:

3 0

2 years ago

Need help again... :(

agasfer [191]

Answer:

Step-by-step explanation:

I think your suppose to find the greatest common factor

5 0

3 years ago