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2 years ago

If two lines intersect in a 90° angle what can you save up lines

Mathematics

1 answer:

Anna007 [38]2 years ago

3 0

If two lines intersect in a 90 degree angle the lines are perpendicular.

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Given the sequence 1/2 ; 4 ; 1/4 ; 7 ; 1/8 ; 10;.. calculate the sum of 50 terms

miv72 [106K]

Hint :-

Break the given sequence into two parts .
Notice the terms at gap of one term beginning from the first term .They are like $\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8}$ . Next term is obtained by multiplying half to the previous term .
Notice the terms beginning from 2nd term , $4,7,10,13$ . Next term is obtained by adding 3 to the previous term .

Solution :- 

We need to find out the sum of 50 terms of the given sequence . After splitting the given sequence ,

$\implies S_1 = \dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8}$ .

We can see that this is in Geometric Progression where 1/2 is the common ratio . Calculating the sum of 25 terms , we have ,

$\implies S_1 = a\dfrac{1-r^n}{1-r} \\\\\implies S_1 = \dfrac{1}{2}\left[ \dfrac{1-\bigg(\dfrac{1}{2}\bigg)^{25}}{1-\dfrac{1}{2}}\right]$

Notice the term $\dfrac{1}{2^{25}}$ will be too small , so we can neglect it and take its approximation as 0 .

$\implies S_1\approx \cancel{ \dfrac{1}{2} } \left[ \dfrac{1-0}{\cancel{\dfrac{1}{2} }}\right]$

$\\\implies \boxed{ S_1 \approx 1 }$

$\rule{200}2$

Now the second sequence is in Arithmetic Progression , with common difference = 3 .

$\implies S_2=\dfrac{n}{2}[2a + (n-1)d]$

Substitute ,

$\implies S_2=\dfrac{25}{2}[2(4) + (25-1)3] =\boxed{ 908}$

Hence sum = 908 + 1 = 909

7 0

2 years ago

What is the area of a triangle with vertices (1,0) (5,0) (3,4)

Aleksandr-060686 [28]

Answer:

The area of the triangle is of 8 units of area.

Step-by-step explanation:

Answer:

The area of the triangle is of 21 units of area.

Step-by-step explanation:

The area of a triangle with three vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by the determinant of the following matrix:

$A = \pm 0.5 \left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\1_3&y_3&1\end{array}\right|$

In this question:

Vertices (1,0) (5,0) (3,4). So

$A = \pm 0.5 \left|\begin{array}{ccc}1&0&1\\5&0&1\\3&4&1\end{array}\right|$

$A = \pm 0.5(1*0*1+0*1*3+1*5*4-1*0*3-0*5*1-1*1*4)$

$A = \pm 0.5*(20-4)$

$A = \pm 0.5*16 = 8$

The area of the triangle is of 8 units of area.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=-3%2Fz%2B7%2F4z%3D5%2Fz-25" id="TexFormula1" title="-3/z+7/4z=5/z-25" alt="-3/z+7/4z=5/z-25" a

ycow [4]

Answer:

z = 1/4

Step-by-step explanation:

See attached image

4 0

2 years ago

Read 2 more answers

How can I get -1 with theses answers

LiRa [457]

You can't get negative 1 from any equation shown in this picture. By the way, -5 +(-9) is not negative 4, it is negative 14,because according to PEMDAS( or BODMAS/BIDMAS/BEDMAS ), parenthesis goes first, so it's kinda like
(-9) + -5 instead.

7 0

3 years ago

Quadratic functions g and k are showed below

padilas [110]

C would equal -3 because you are going up 9 units

6 0

3 years ago