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insens350 [35]
4 years ago
13

Determine whether the given measures can be the lengths of the sides of a triangle write yes or no

Mathematics
1 answer:
ohaa [14]4 years ago
3 0

NO.,the given measures can not be the lengths of the sides of a triangle

Step-by-step explanation

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

so, Find the range for the measure of the third side of a triangle given the measures of two sides.

here given measures are 2,2,6

2+2 = 4 which is less than the third side 6

        = 4 < 6    

This not at all a triangle.

Hence, the given measures can not be the lengths of the sides of a triangle

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A right rectangular prism's edge lengths are 14 inches, 6 inches, and 3 1/3 inches. How many unit cubes with edge lengths of 2/5
DochEvi [55]

ANSWER

4375 unit cubes

EXPLANATION

The volume of a right rectangular prism is given by;

V=lbh

We substitute the given dimension to obtain,

V=14 \times 6 \times  \frac{10}{3}  = 280i {n}^{3}

The volume of the unit cube with edge lengths

\frac{2}{5}

is

(\frac{2}{5})^{3}  =  \frac{8}{125}  {in}^{3}

To find the number of unit cubes that can fit inside the prism, we divide the volume of the rectangular prism by the volume of the unit cube.

=  \frac{280}{ \frac{8}{125} }

= 4375

7 0
3 years ago
How do you do this question?
Lena [83]

Step-by-step explanation:

The Taylor series expansion is:

Tₙ(x) = ∑ f⁽ⁿ⁾(a) (x − a)ⁿ / n!

f(x) = 1/x, a = 4, and n = 3.

First, find the derivatives.

f⁽⁰⁾(4) = 1/4

f⁽¹⁾(4) = -1/(4)² = -1/16

f⁽²⁾(4) = 2/(4)³ = 1/32

f⁽³⁾(4) = -6/(4)⁴ = -3/128

Therefore:

T₃(x) = 1/4 (x − 4)⁰ / 0! − 1/16 (x − 4)¹ / 1! + 1/32 (x − 4)² / 2! − 3/128 (x − 4)³ / 3!

T₃(x) = 1/4 − 1/16 (x − 4) + 1/64 (x − 4)² − 1/256 (x − 4)³

f(x) = 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0.  So we can eliminate the top left option.  That leaves the other three options, where f(x) is the blue line.

Now we have to determine which green line is T₃(x).  The simplest way is to notice that f(x) and T₃(x) intersect at x=4 (which makes sense, since T₃(x) is the Taylor series centered at x=4).

The bottom right graph is the only correct option.

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