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3 years ago

Which of the following similarity statements about the triangles in the figure is true?

Mathematics

1 answer:

maksim [4K]3 years ago

7 0

C is correct. You have to pay attention to the corresponding sequences of the vertices. C is correct because in the three triangles, MON, MPO, OPN are the right angles, respectively. However, the angles are not consistent in other given choices.

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Estimate how far he ran in 4 seconds. and Is Kiran's time a function of the distance he has run?

AysviL [449]

Answer:

1.5 for how far he ran in 4 seconds

Step-by-step explanation:

3 0

2 years ago

Is -8 - 2(3+2n)+7n equivalent to -30 - 13n

soldi70 [24.7K]

Answer:

The terms are not equivalent

Step-by-step explanation:

-8 - 2(3+2n)+7n

Distribute the 2

-8 -6-4n+7n

Combine like terms

-14 +3n

That is not equal to -30 -13n

The terms are not equivalent

4 0

3 years ago

An opaque bag contains 5 green marbles 3 blue marbles and 2 red marbles. What is the probability of randomly picking out a marbl

anastassius [24]

Answer= .2 probability

Explanation: Since there is a total of 10 marbles, and 2 are red, you would divide the red marbles (2) out of the total marbles (10) which would get u .2. Since it’s asking for probability, we would keep .2 a decimal.

6 0

2 years ago

Prove the trigonometric identity

Annette [7]

Answer:

Proved See below

Step-by-step explanation:

Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

$1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1$

Lets solve the bottom most part first:

$1-\frac{1}{1-sec^2x} \\\\$

Take LCM

$1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}$

now break the LCM

$\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1$

because 1/tan = cot x

and furthermore,

$cot^2x+1\\cosec^2x$

now we solve the above part and replace the bottom most part that we solved with $cosec^2x$

$\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x$

Hence proved! :D

4 0

2 years ago

Use the formula to evaluate the series 1/4 + 1/9 + 1/27 + 1/81.

Dmitriy789 [7]

Is the first term supposed to be 1/4, or did you actually mean 1/3?

I'll assume the latter. Write

$S=\dfrac13+\dfrac19+\dfrac1{27}+\dfrac1{81}$
$S=\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}$
$\dfrac13S=\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}+\dfrac1{3^5}$
$\implies S-\dfrac13S=\dfrac13-\dfrac1{3^5}$
$\dfrac23S=\dfrac13\left(1-\dfrac1{3^4}\right)$
$S=\dfrac12\left(1-\dfrac1{81}\right)$
$S=\dfrac{40}{81}$

8 0

3 years ago