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3 years ago

In ∆ABC, point D is the centroid. If (AE) ̅=15, (BF) ̅=27, and (CD) ̅=8, find the following lengths

Mathematics

1 answer:

Anna71 [15]3 years ago

7 0

Answer:

5,18,12 cms are the answer.

Step-by-step explanation:

Given is a triangle ABC. Point D is the centroid.

E,F and G are midpoints of CB, BA and AC respectively.

AE, BF and CG are medians of the triangle.

We know that centroid divides the median in the ratio 2:1

Using this we find that AD:DE = 2:1

Or AD+DE:DE = (2+1):1

AE:DE =3:1

15:DE = 3:1 . Hence DE =5 cm.

On the similar grounds we find that DF = 1/3 BF = 9

Hence BD = DF-BF = 27-9 =18 cm

and also

CG = 3/2 times CD = 12 cm.

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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

$19900 - 4900 - 2842 + 784 = 12942$ .

8 0

3 years ago

What is the factor of 28/88

Llana [10]

The factors are:
4, 2, 1

6 0

3 years ago

How many different sums can be made with a penny nickel dime quarter?

weqwewe [10]

1 coin is 4 ways
2 coins is 4*3 ways
3 coins is 4*3*2 ways
4 coins is 4*3*2*1 ways

so total is 4+12+24+24=64 sums

8 0

3 years ago

Need help ! Simplify the rational expression.state any excluded values.2x/-5x+x^2

mario62 [17]

I assume it is:

2x/(-5x+x^2), then you can simplify the x, but only if it is not zero:

2/(x-5), for x different to zero.

3 0

3 years ago

A company plans to build at least 152 igloos in a single week by employing skilled workers from the cities of Tomsk and Kyzyl. A

xxTIMURxx [149]

Answer:

Each Tomskian build 6 igloos per week.

Each Kyzylian build 5 igloos per week.

Step-by-step explanation:

Let T represent the number of Tomskians and K represent the number of Kyzylians the company needs to employ to meet its goal.

The inequality: $6T+5K\geq152$ represents the situation that a company plans to build at least 152 igloos in a single week by employing T Tomskians and K Kyzylians.

We can see that 6T represents the number of igloos made by T Tomskians in one week and 5K represents the number of igloos made by K Kyzylians in one week .

Since all Tomskians and Kyzylians build the same number of igloos per week, therefore, each Tomskian build 6 igloos per week and each Kyzylian build 5 igloos per week.

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3 years ago

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