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3 years ago

11 points please help and give a small explanation I’ll mark brainliest

Mathematics

2 answers:

faltersainse [42]3 years ago

7 0

Answer:

what are we looking at

Step-by-step explanation:

dalvyx [7]3 years ago

4 0

Answer:

31.5 IS YOUR ANSWER

2in=7m. 3in=10 1/2. 4in=14m. 6in=21m. 8in=28. 9in=31.5

TO GET THE METERS ADD 31/2 WHICH IS HALF OF 7.

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Find the value of y in the diagram below.

Alja [10]

Answer:

The answer to your question is: y = 80°

Step-by-step explanation:

The sum of the internal angles in a polygon is = 180°(n - 2)

= 180° (6 - 2)

= 180(4)

= 720°

y° + y° + 2(2y -20) + 2(2y -20) = 720

2y° + 4(2y -20) = 720

2y° + 8y -80 = 720

10y = 720 + 80

10 y = 800

y = 800 / 10

y = 80°

2y - 20 = 2(80) -20 = 160 -20 = 140

80° + 80° * 140 + 140 + 140 + 140 = 720°

5 0

3 years ago

3. Anne is playing a math game. She chooses three

Anastaziya [24]

Answer: Im pretty sure the answer is -14

Step-by-step explanation:

6 0

3 years ago

Label each statement as true or false regarding the zeros/roots of a quadratic function. The roots zeros of a quadratic function

ikadub [295]

Answer:

True, false, true, true.

Step-by-step explanation:

The roots zeros of a quadratic function are the same as the factors of the quadratic function. This is true because your roots are your factors—>(x-3) is a factor, x=3 is the root.

The roots zeros are the spots where the quadratic function intersects with the y-axis. No! Those are called y-intercepts!

The roots zeros are the spots where the quadratic function intersects with the x-axis. True. X-intercepts are your solutions. (x-3) graphed would the (3,0). That’s a solution.

There are not always two roots/zeros of a quadratic function, True. No solution would be when your quadratic doesn’t intersect the x-axis. One solution would be when your vertex would be on the x-axis. Two solutions is when your quadratic intersects the x-axis twice. Can there be infinite solutions? No. It’s either 0, 1, or 2 solutions.

6 0

3 years ago

Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]

Schach [20]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

$\rm \: = \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

$\rm \: = \: \dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }$

Now, Consider Denominator, we have

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

Now, we know that,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}$

So, using this, we get

$\rm \: = \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

$\rm \: = \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \: = \: \dfrac{e - 1}{e}$

$\rm \: = \: 1 - \dfrac{1}{e}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} } = \frac{e - 1}{e} = 1 - \frac{1}{e}}}$

3 0

3 years ago

Simplify (y x y^1/3)^3/2

jok3333 [9.3K]

Answer:

y^2

Step-by-step explanation:

(y x y^1/3)^3/2 would be clearer if written as

(y*y^(1/3))^(3/2).

Because y^a*y^b = y^(a + b), the quantity inside the first set of parentheses becomes

y^(4/3)

and if we apply the power (3/2) to this result, we get:

4 3

(y^(4/3))^(3/2) = y^{ ----- * ----- ) = y^2

3 2

3 0

3 years ago