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3 years ago

PLEASE ANSWER

Mathematics

2 answers:

Nataliya [291]3 years ago

7 0

Answer: it’s the 2nd, 3rd and 5th one

vekshin13 years ago

4 0

Answer:

Step-by-step explanation:

Given are some properties of triangles and we have to check whether they are correct

i) Only two of three angle bisectors of the internal angles of a triangle are concurrent.

This is incorrect since all three angle bisectors concur at incentre

ii) The circumcenter of a triangle is the point where the perpendicular bisectors of the sides meet.

-- correct because the meeting point is equidistant from all three vertices

iii) Given any three non-collinear points, there exists exactly one circle that passes through the points.

-- correct because any three points determine a circle

iv) Given any three non-collinear points, there exists exactly one circle that passes through the points.

-- Correct

v) The incenter of a triangle is the point where the angle bisectors meet.

-- Correct and the centre is equidistant from the sides of the triangle

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23.

Crazy boy [7]

Answer:

The number that cannot be the largest possible 6-digit number is;

(D) AAABCB

Step-by-step explanation:

From the question, we have;

A, B, and C = Distinct digits, therefore, A ≠ B ≠ C

The number of digits in the number to be formed = 6 digits

The number of 'A' in the number to be formed = 3

The number of 'B' in the number to be formed = 2

The number of 'C' in the number to be formed = 1

We have;

When A > B > C

The largest possible number = AAABBC

When C > A > B

The largest possible number = CAAABB

When B > A > C

The largest possible number = BBAAAC

When A > C > B

The largest possible number = AAACBB

Therefore, given that when A > B > C, the largest possible number = AAABBC, we have;

AAABBC > AAABCB, because B > C, therefore, within the tens and unit of the two 6 digit numbers, we have, BC > CB

∴ AAABBC > AAABCB and AAABCB, cannot be the largest possible 6-digit number

3 0

3 years ago

Solve the equation. 7+6=-3y+26 A. y= -8 B. y= -5 C. y= 5 D. y= 8

Umnica [9.8K]

The correct answer is c y=5

7 0

3 years ago

A city manager made a scatter plot of the number or retail store in a certain city over the years. the scatter plot had a trend

Fiesta28 [93]

Answer:

155

Step-by-step explanation:

The problem sounds complicated, but it's not. Let's analyse.

The city manager have come up with an equartion y=11x +12, with Y is the total number of the stores and X stands for how long it has been since 2003.

We can't explain how the manager came up with this equation, so iwe don't need to think of if the equation is real or not. Let's just base on what we have.

Because the equation aboce is a trendy line, it means that it would likely to be true with any X ( number of years since 1990).

In Tracy's case, the year is 2003, so it has been 2003 - 1990 = 13 years since 1990. This is the X in the equation. Now we only need to find Y in the equation, which is the number of retail stores there were in 2003, exactly what the problem asks.

y= 11x + 12

=> In Tracy's case: y= 11*13 + 12= 155

So the number of retail stores there were in 2003 was 155

5 0

3 years ago

To help divide a group into 4 teams, the teacher brought out a bag with equal numbers of 4 different colored popsicle sticks (gr

NeX [460]

The set of all possible events Ω

Ω = 24 ( 4*7 = 28 stick)

set of events favorable A
A = 7 ( sticks of green is 7)

Probability P
P(A) = A/Ω = 7/28 = 1/4 = 0,25

Answer A
The first person has the ability to draw seven green sticks of twenty-four

3 0

3 years ago

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

5 0

3 years ago