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ikadub [295]
3 years ago
9

Determine if the two triangles shown are similar. If so, write the similarity statement.

Mathematics
1 answer:
Ludmilka [50]3 years ago
5 0

Answer:

C. The triangles are not similar

Step-by-step explanation:

For two triangles to be considered similar, the 3 angles of one triangle must be equal or congruent to the corresponding 3 angles of the others.

The only angle in ∆BCG that is equal to the corresponding angles in ∆EFG is <BGC. The remaining 2 angles in ∆BCG are not congruent to the corresponding 2 angles in ∆EFG.

Therefore, we can conclude that both triangles are not similar.

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A. y = (X - 10)^3 - 12
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PLEASE ANSWER ASAP!!! Using the given equation find the missing coordinates of the points and then find the slope of the line fo
Andrei [34K]

Answer:

Step-by-step explanation:A.

B.

Slope =

Step-by-step explanation:

We are given the equation of the line as .

A. The co-ordinate is given by

Substituting the value , we get,

implies  implies   i.e. 3x = -1 i.e.

Thus, the co-ordinate is .

B. The co-ordinate is given by

Substituting the value , we get,

implies  implies   i.e. 9y = 0 i.e. y= 0

Thus, the co-ordinate is .

Since, the slope of a line is given by .

We get,

Slope = .

i.e. Slope = .

i.e. Slope = .

Hence, the slope of the line is .

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Please help!!! ASAP the bottom number is 3 btw.
Nataliya [291]

Answer:

5*3=15

Step-by-step explanation:

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3 years ago
Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t
Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1.f(x)=\frac{x^2+x-20}{x^2+4}

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345

3.h(x)=\frac{3x-5}{x^2-5x+7}

x^2-5x+7=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43

2.g(x)=\frac{x-17}{x^2+75}

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12

4.i(x)=\frac{x^2-9}{x-9}

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is  not continuous on the set of real numbers.

5.j(x)=\frac{4x^2-7x-65}{x^2+10}

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0

6.k(x)=\frac{x+1}{x^2+x+29}

x^2+x+29=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102

7.l(x)=\frac{5x-1}{x^2-9x+8}

x^2-9x+8=0

x^2-8x-x+8=0

x(x-8)-1(x-8)=0

(x-8)(x-1)=0

x=8,1

The function is not defined for x=8 and x=1

Hence, function l is not  defined for all real values.

8.m(x)=\frac{x^2+5x-24}{x^2+11}

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

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Ilia_Sergeevich [38]

a. The \vec k component tells you the particle's height:

3t=16\implies t=\dfrac{16}3

b. The particle's velocity is obtained by differentiating its position function:

\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+3\,\vec k

so that its velocity at time t=\frac{16}3 is

\vec v\left(\dfrac{16}3\right)=-\sin\dfrac{16}3\,\vec\imath+\cos\dfrac{16}3\,\vec\jmath+3\,\vec k

c. The tangent to \vec r(t) at t=\frac{16}3 is

\vec T(t)=\vec r\left(\dfrac{16}3\right)+\vec v\left(\dfrac{16}3\right)t

4 0
3 years ago
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