BRAINLY!!!- if a=7, b=8, and c=9.2 is this a right triangle? why?

Mathematics

1 answer:

Travka [436]3 years ago

7 0

Answer:

Step-by-step explanation:

If this were a right triangle we would assume the Pythagorean Theorem to prove it

$a^{2} +b^{2} =c^{2}$

$7^{2} +8^{2} =9.2^{2}$

this gives us

49+64=84.64

this is 143=84.64 and therefore not equal and not a right triangle

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Xelga [282]

Answer:

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Step-by-step explanation:

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3 years ago

Solve the triangle A = 2 B = 9 C =8

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

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1 year ago

Which equations may be used to determine the 3 consecutive integers whose sum is 127?

s344n2d4d5 [400]

Answer:

a and b.

Step-by-step explanation:

I’ll explain by giving an example.

Let’s say that: a=3;b=4;c=5; => they all are consecutive -> their sum is 12.

=> if we use a) n=3 => 3*n+3=3*3+3=12 => correct.

b) n+(n+1)+(n+2)= 3+4+5=12=> correct.

c)n+2n+3n=3+6+9=18=>incorrect.