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3 years ago

Please help. thank you.

Mathematics

2 answers:

ruslelena [56]3 years ago

7 0

Answer: 150(1 + 4) 16

Step-by-step explanation: 4 goes in the first box and 16 goes in the little one.

Tresset [83]3 years ago

3 0

Usuageieisbahisisvshriirjdvsvnzkvog

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The set M represents multiples of 4 from 6 to 38. M=8,12,16,20,24,28,32,36 The set N represents even numbers from 4 to 13. N=4,6

Tomtit [17]

Answer:

The correct option is B. 2

So there are two elements that is 8, 12 in the set M⋂N .

Step-by-step explanation:

Given:

M is a set of multiples of 4 from 6 to 38

i.e M = { 8, 12, 16, 20, 24, 28, 32,36 }

N is a set of even numbers from 4 to 13

i.e N = { 4, 6, 8, 10, 12 }

To Find:

M⋂N = ?

Solution:

Set:

A set is a well-defined collection of distinct objects, considered as an object in its own right.

Examples :

M is a set of multiples of 4 from 6 to 38

i.e M = { 8, 12, 16, 20, 24, 28, 32,36 }

N is a set of even numbers from 4 to 13

i.e N = { 4, 6, 8, 10, 12 }

M⋂N = set of those numbers which are present in M as well as in N

∴ M⋂N = { 8, 12 }

So there are two elements that is 8, 12 in the set M⋂N .

8 0

3 years ago

. A right triangle has a base of 6 centimeters and the height is

Evgen [1.6K]

So did have eieheuehw die is ruejejd ri

5 0

3 years ago

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I really stuck on trying to prove this

Alisiya [41]

$\displaystyle\sum_{r=1}^nr(r+1)\cdots(r+p-1)$

When $n=1$ ,

$\displaystyle\sum_{r=1}^1r(r+1)\cdots(r+p-1)=1(1+1)(1+2)\cdots(1+p-2)(1+p-1)=p!$

Meanwhile, you have on the right

$\dfrac{(1)(1+1)(1+2)\cdots(1+p-2)(1+p-1)(1+p)}{p+1}=(1)(1+1)(1+2)\cdots(p-1)(p)=p!$

so the equality holds for $n=1$ .

Assume it holds for $n=k$ , i.e. that

$\displaystyle\sum_{r=1}^kr(r+1)\cdots(r+p-1)=\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}$

Now for $n=k+1$ , you have

$\displaystyle\sum_{r=1}^{k+1}r(r+1)\cdots(r+p-1)=\sum_{r=1}^kr(r+1)\cdots(r+p-1)+(k+1)(k+2)\cdots(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+1+p-2)(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\left(\dfrac k{p+1}+1\right)(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{k+p+1}{p+1}(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{(k+1)(k+2)\cdots(k+p-1)(k+p)(k+p+1)}{p+1}$

as required.

3 0

3 years ago

Solve the system of equations:<br> y=2x<br> Y=x^2-3

ohaa [14]

Answer:

Step-by-step explanation:

The two solutions are (-1,-2) and (3,6)

5 0

3 years ago

In the diagram, which must be true for point D to be an orthocenter?

Misha Larkins [42]

A copy of the diagram is shown below

Point D is the intersection of three angle bisector.
BE is the angle bisector of ∠B
CF is the angle bisector of ∠C
AG is the angle bisector of ∠A

Point D is also the intersection between three perpendicular bisector
BE is the perpendicular bisector of AC
AG is the perpendicular bisector of BC
CF is the perpendicular bisector of AB

Hence the correct statements is statement 1 and statement 4

6 0

4 years ago

Read 2 more answers