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3 years ago

50 POINTS!

Mathematics

2 answers:

Hoochie [10]3 years ago

7 0

Answer:

Solution given:

AB=3units

<B=45°

<C=63°

DF=3units

<E=63°

<F=45°

for congruence opposite corresponding angle or side should be equal

∆ABCis congruent to ∆DFE by A.A.S axiom

Similarly

its another name also be

∆BAC=~ ∆FDE

you got it.third one.

erik [133]3 years ago

7 0

Answer:

C (could be D but I think C)

Step-by-step explanation:

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Andru [333]

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$\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}$

If you're having problems understanding the answer, try seeing it through your browser: brainly.com/question/2112248

$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}$

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How do i solve this?

Damm [24]

From the ratio given you can determine that m is proportional to 1/(r^3). You can write this as m = k * 1/(r^3), k being your constant of proportionality. Then use this to calculate the next parts of the question.

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