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2 years ago

25. Given: TE= RI, TI = RE, Proof LTDI and LROE are right / Prove: TD=RO

Mathematics

1 answer:

zaharov [31]2 years ago

5 0

Answer:

In triangle TEI and ERI

TI = RE ( given)

angle TDI = angle ROE ( given).

TE = RI ( given).

by side angle side criteria

both the triangles are congruent

by cpct ( corresponding parts of congruent triangles)

TD is congruent to RO

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Answer:

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Step-by-step explanation:

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4 - 5/2 * (1/10x) = 1 what is the value of x

Inga [223]

$4 - 5 / 2 * (1/10x) = 1\\$

The first step is to identify the order in which the equation must be solved, by following PEMDAS (you might know it as BEDMAS):

Parenthesis (or Brackets)

Exponents

Multiplication and Division

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My advice would be to add parenthesis, following these rules, if you are not very good at finding them immediately by sight.

So:

$4 - 5 / 2 * (\frac{1}{10x}) = 1\\\\4 - [(5/2)*(\frac{1}{10x})]=1\\\\4-(2.5*\frac{1}{10x})=1\\\\4-\frac{2.5}{10x}-1=0\\3-\frac{x}{4}=0\\\frac{x}{4}=3\\x=3*4\\x=12$

We check our answer:

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3 years ago

A school director must randomly select 6 teachers to part in a training session. There are 34 teachers at school. In how many di

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Answer:

Total possible ways to select 6 teachers from 34 teacher are $^{34}C_6=1344904$ .

Step-by-step explanation:

It is given that total number of teachers at a school is 34.

The school director must randomly select 6 teachers to part in a training session.

$^nC_r=\frac{n!}{r!(n-r)!}$

Where, n is total possible outcomes and r is number of selected outcomes.

Total teachers = 34

Selected teachers =6

Total number of possible ways to select 6 teachers from 34 teacher is

$^{34}C_6=\frac{34!}{6!(34-6)!}$

$^{34}C_6=\frac{34\times 33\times 32\times 31\times 29\times 28!}{6!(28)!}$

$^{34}C_6=1344904$

Therefore total possible ways to select 6 teachers from 34 teacher are $^{34}C_6=1344904$ .

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Elanso [62]

Answer:

Choose the first alternative

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Step-by-step explanation:

Probabilities

The requested probability can be computed as the ratio between the number of ways to choose two sophomores in alternate positions $(N_s)$ and the total number of possible choices $(N_t)$ , i.e.

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There are 6 sophomores and 14 freshmen to choose from each separate set. There are 20 students in total

We'll assume the positions of the selections are NOT significative, i.e. student A/student B is the same as student B/student A.

To choose 2 sophomores out of the 6 available, the first position has 6 elements to choose from, the second has now only 5

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The total number of possible choices is

$_{2}^{20}\textrm{C} \text{ ways to do it}$

The probability is then

$\boxed{\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}}$

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