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4 years ago

In an isosceles triangle the base angles are congruent. True False

Mathematics

2 answers:

Alisiya [41]4 years ago

7 0

Hi, it will be true.

I hoped I helped and if you need anymore you could always ask me :)

-Dawn

kherson [118]4 years ago

4 0

This statement is false. Because the base angles of an isosceles triangle arecongruent, if one base angle is a right angle then both base angles must be rightangles. It is impossible to have a triangle with two right (90^\circ)angles.

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Simplify 3^2 x 3^4 x 3^6 A3^10 B3^12 C3^48 D3^0

kvasek [131]

Answer:

$\boxed{\sf\: B)\:3^{12}}$

Step-by-step explanation:

$\sf 3^2\times \:3^4\times \:3^6$

Apply exponent rule:

$\hookrightarrow \boxed{\sf a^b\times \:a^c=a^{b+c}}$

$\sf 3^2\times \:3^4\times \:3^6$

$\sf 3^{2+4+6}$

Add the numbers:

$\sf 2+4+6=12$

$\sf \:3^{12}$

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2 years ago

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Vanyuwa [196]

The second:
$\left \{ {{-2x+4y=18} \atop {4x-8y=-36}} \right.$
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3 years ago

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makkiz [27]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

We used Total - 1 because it is a probability without replacement

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$