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1 year ago

The 9th and the 12th term of an arithmetic progression are 50 and 65 respectively. find the common difference

Mathematics

1 answer:

Rzqust [24]1 year ago

4 0

The first term of the arithmetic progression exists at 10 and the common difference is 2.

<h3>How to estimate the common difference of an arithmetic progression?</h3>

let the nth term be named x, and the value of the term y, then there exists a function y = ax + b this formula exists also utilized for straight lines.

We just require a and b. we already got two data points. we can just plug the known x/y pairs into the formula

The 9th and the 12th term of an arithmetic progression exist at 50 and 65 respectively.

9th term = 50

a + 8d = 50 ...............(1)

12th term = 65

a + 11d = 65 ...............(2)

subtract them, (2) - (1), we get

3d = 15

d = 5

If a + 8d = 50 then substitute the value of d = 5, we get

a + 8 $*$ 5 = 50

a + 40 = 50

a = 50 - 40

a = 10.

Therefore, the first term is 10 and the common difference is 2.

To learn more about common differences refer to:

brainly.com/question/1486233

#SPJ4

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I am confused please help

Otrada [13]

Answer:

3 tiles will not fit together.

Step-by-step explanation:

Measure of an Interior angle of a polygon = $\frac{(n-2)\times 180}{n}$

Here, n = number of sides of the polygon

Therefore, measure of the interior angles of a regular hexagon,

A = $\frac{(6-2)\times 180}{6}$

A = 120°

Similarly, interior angle of the regular pentagon,

B = $\frac{(5-2)\times 180}{5}$

B = 108°

Now m∠A + m∠B + m∠C = 360°

m∠C = 360° - (120° + 108°)

= 132°

To fit the given three tiles perfectly, interior angle (∠D) of the third Octagonal tile should be 132°.

D = $\frac{(8-2)\times 180}{8}$

D = 135°

m∠C ≠ m∠D

Therefore, 3 tiles will not fit together.

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2 years ago

A fair coin is tossed four times. What is the probability that the number of heads appearing on the first two tosses is equal to

alekssr [168]

Answer: 3/8

Step-by-step explanation:

Since it is a fair coin, then generally, P(Head) = P(Tail) = ½

And since we've been asked to find the probability that the number of heads in the first two tosses be equal to the number of heads in the second two tosses, tossing a fair coin four times, the possible outcomes of having equal number of heads in first two tosses and second two tosses becomes:

[HHHH] or [HTHT] or [THTH] or [TTTT] or [HTTH] or [THHT]

=[½*½×½*½] + [½*½×½*½] + [½*½×½*½] + [½*½×½*½] + [½*½×½*½] + [½*½×½*½]

=1/16 * 6

=6/16

=3/8.

8 0

2 years ago

cora says she doesn't need to know the y-intercept of a line to write its equation, just its slope and some other point on the l

Nadusha1986 [10]

Answer:

The equation of line with given slope that include given points is 3 y + x - 20 = 0

Step-by-step explanation:

According to Cora , if we know the slope and points on a line then we can write the equation of a line .

Since , The equation of line in slope-intercept form is

y = m x + c

<u>Where m is the slope of line , and if we know the points ( x , y ) which satisfy the line then constant term c can be get and the equation of line can be formed .</u>

So , From the statement said above it is clear that she is correct .

Now , Again

Given as :

Slope of a line is m = - $\frac{1}{3}$

That include points ( 2 , 6 )

Now from the equation of line as y = m x + c

∴ 6 = - $\frac{1}{3}$ ( 2 ) + c

Or, 6 = - $\frac{2}{3}$ + c

So , c = 6 + $\frac{2}{3}$

or, c = $\frac{18 + 2}{3}$

∴ c = $\frac{20}{3}$

So, The equation of line can be written as

y = - $\frac{1}{3}$ x + $\frac{20}{3}$

Or, 3 y = - x + 20

I.e 3 y + x - 20 = 0

Hence The equation of line with given slope that include given points is 3 y + x - 20 = 0 Answer

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2 years ago

In the graphic below, ARV = ENV by:

GREYUIT [131]

The ans should be ASA because angle AVR is equal to angle EVN (opposite angles equal)

8 0

3 years ago

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of <em>x</em> ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As <em>x</em> goes to infinity, each of the 1/<em>x</em> ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

2 years ago