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4 years ago

Which of the following side lengths can form a triangle?

Mathematics

2 answers:

Dmitry [639]4 years ago

5 0

Answer:

side lengths can form a triangle

21 cm, 41 cm, and 60 cm..

Pachacha [2.7K]4 years ago

3 0

I'll be solving this by using the Triangle Inequality Theorem: a+b>c, a+c>b, and b+c>a

A. 54 mm, 65 mm, and 120 mm: 54=a, 65=b, 120=c and 54 + 65 > 120 so it's false.

B. 12 in, 43 in, and 56 in: 12=a, 43=b, 56=c, 12 + 43 > 56 so false

C. 16 ft, 24 ft, and 42 ft: 24=a, 16=b, 42=c, but 16 + 24 > 60 so it's false

D. 21 cm, 41 cm, and 60 cm: 21=a, 41=b, 60=c

21 + 41 > 60

21 + 60 > 41

41 + 60 > 21

It equals all three formulas so D is the right answer.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Valentines Day! :-)

- Cutiepatutie ☺❀❤

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3 years ago

A hiker is hiking in a valley. The height of the valley is h(x,y)=4x2+y2 where x and y are the east-west and north-south distanc

Ainat [17]

Answer:

A. $\frac{\partial{h}}{\partial{t}}=0$

Step-by-step explanation:

A. The problems asked for 2 ways to solve it, expanding the equation with the substitution x(t)=2 cos(t) and y(t)=4 sin(t) to differentiate it . The other way is by chain rule.

Expanding and differentiating:

We start by substituting x(t)=2 cos(t) and y(t)=4 sin(t) in h(x,y)=4x2+y2:

$h(x,y)=4x^{2}+y^{2}= 4(2cos(t))^{2}+(4sin(t))^{2}\\h(x,y)=4(4cos^{2}(t))+(16sen^{2}(t))\\h(x,y)=16cos^{2}(t)+16sen^{2}(t)=16(sen^{2}(t)+cos^{2}(t))\\h(x,y)=16$

So, in the path that the hiker chose:

$\frac{\partial{h}}{\partial{t}}=0$

Chain rule:

We start differentiating h(x,y) using chain rule as follows:

$\frac{\partial{h}}{\partial{t}}= \frac{\partial{h}}{\partial{x}}\frac{\partial{x}}{\partial{t}}+\frac{\partial{h}}{\partial{y}}\frac{\partial{y}}{\partial{t}}$

Now, it´s easy to find all these derivatives:

$\frac{\partial{h}}{\partial{x}}=8x\\\frac{\partial{x}}{\partial{t}}=-2sin(t)\\\frac{\partial{h}}{\partial{y}}=2y\\\frac{\partial{y}}{\partial{t}}=4cos(t)$

Now we replace them in the chain rule, with the replacement x=2cos(t) and y=4sin(t) in the x,y that are left and we operate everything:

$\frac{\partial{h}}{\partial{t}}= 8x(-2sin(t))+2y(4cos(t)$

$\frac{\partial{h}}{\partial{t}}= 8(2cos(t))(-2sin(t))+2(4sin(t))(4cos(t)$

$\frac{\partial{h}}{\partial{t}}= -32cos(t)sin(t)+32sin(t)cos(t)$

$\frac{\partial{h}}{\partial{t}}= 0$

This will be our answer

6 0

3 years ago

Which ordered pairs make both inequalities true? Check all that apply.

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All others did not satisfy the inequality simultaneously except #4 and #6
For #4:
2 < 5(1) + 2 i.e. 2 < 7 (true)
2 >= 1/2(1) + 1 i.e. 2 >= 3/2 (true)
For #6
2 < 5(2) + 2 i.e. 2 < 12 (true)
2 >= 1/2(2) + 1 i.e. 2 >= 2

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3 years ago

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