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Katyanochek1 [597]
3 years ago
14

What would prove that the two triangles are congruent?

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

The answer must be R is approximately equal to C because as you can see on the picture, Triangle AKR is upside down from MYC, and R is the pointy edge on the AKR triangle, and C is the pointy edge on the MYC triangle.

Step-by-step explanation:

aniked [119]3 years ago
5 0
The lines on the sides!
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what is the quotient (65y3 15y2 − 25y) ÷ 5y? a. 13y2 3y − 5 b. 13y3 3y2 − 5y c. 13y2 − 3y 5 d.13y2 − 3y − 5
Aneli [31]

Keywords:

<em>Division, quotient, polynomial, monomial </em>

For this case we must solve a division between a polynomial and a monomial and indicate which is the quotient.

By definition, if we have a division of the form: \frac {a} {b} = c, the quotient is given by "c".

We have the following polynomial:

65y ^ 3 + 15y ^ 2 - 25y that must be divided between monomy5y, then:

C (y) represents the quotient of the division:

C (y) = \frac {65y ^ 3 + 15y ^ 2 - 25y} {5y}

C (y) = \frac {65y ^ 3} {5y} + \frac {15y ^ 2} {5y} - \frac {25y} {5y}

C (y) = 13y ^ 2 + 3y-5

Thus, the quotient of the division between the polynomial and the monomial is given by:

C (y) = 13y ^ 2 + 3y-5

Answer:

The quotient is: C (y) = 13y ^ 2 + 3y-5

Option: A


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2 years ago
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What is the range of 1/2 1/8 1/2 5/8 1/4 1/4 1/2 3/4 1/8 5/8 1/4 1/2
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Answer:

it is 1/8- 5/8 smallest to biggesst

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Which of these is a correct statement?
Montano1993 [528]

Answer:

the answer is B because it has only one unknown and also it is a linear equation so it will have only one answer which is (2 )

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How to factor out 7x^6
Nataly_w [17]

Answer:

42

Step-by-step explanation:

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Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}

a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
11 months ago
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