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3 years ago

Match the given slope (m) and y-intercept (b) with the equation of the line in slope-intercept form.

Mathematics

1 answer:

Allisa [31]3 years ago

7 0

Answer:

y=2x+7

y=4x+7

y=5x

Step-by-step explanation:

substitute the numbers for the variables in the form y=mx+b

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1. What is the value of x? Show your work to justify your answer.

Reil [10]

Answer:

x=72 and the exterior angle is 154

Step-by-step explanation:

We will call the unknown angle in the triangle y. Angle y and the angle (2x +10) form a straight line so they make 180 degrees.

y + 2x+10 =180

Solve for y by subtracting 2x+10 from each side.

y + 2x+10 - (2x+10) =180 - (2x+10)

y = 180-2x-10

y = 170-2x

The three angles of a triangle add to 180 degrees

x+ y+ 82 = 180

x+ (170-2x)+82 = 180

Combine like terms

-x +252=180

Subtract 252 from each side

-x+252-252 = 180-252

-x = -72

Multiply each side by -1

-1*-x = -72*-1

x=72

The exterior angle is 2x+10. Substitute x=72 into the equation.

2(72)+10

144+10

154

8 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

2 years ago

Given f(x) = x+2 and g(x) = x-4 solve for (fg)(x)

kirza4 [7]

Answer: f(g(x)) = x - 2

Step-by-step explanation:

f(x) = x + 2

g(x) = x - 4

f(g(x)) = f(x - 4) = (x - 4) + 2 replaced x with x - 4

= x - 2 simplified

6 0

3 years ago

Rai compares two cable plans from different companies. Which equation gives the correct value of m, the number of months for whi

tekilochka [14]

Answer:

D. 75m = 50(12)+125(m-12)

Step-by-step explanation:

m = months.

Equation for which Plans A and B cost the same.

In Plan A:

No initial fees.

$75 per month

1 month = $ 75

then;

for m month =$75 m

Total cost for plan A = 75m

In Plan B:

$50 per month for first 12 month.

1 month = $50

12 months = 50(12)

Similarly,

$125 per month for each additional month after that.

additional month= (m-12)

Plan B additional month

= 125(m-12)

Total cost for plan B = 50(12)+ 125(m-12)

Since, Plans A and B cost the same.

75m = 50(12)+125(m-12)

3 0

3 years ago

What is the circumference of a circular disc with a diameter of 10 centimeters? (Round your answer to the nearest tenth.)

Evgen [1.6K]

To determine what the circumference is simply take the value of the diameter and multiply it to the value of Pi or 3.141....

C = Pi • d
C = (3.14)(10 cm)
C = 31.41 cm or 31.4 cm, rounded to the nearest tenth.

5 0

3 years ago