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2 years ago

What are the first three terms of the sequence defined by the following function?

Mathematics

1 answer:

Naily [24]2 years ago

7 0

Answer:

Step-by-step explanation:

$f(n)=41+(n-1)^5\\\\f(1)=41+(1-1)^5=41\\\\f(2)=41+(2-1)^5=41-1=40\\\\f(3)=41+(3-1)^5=41-32=9$

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A. Sin40=6/x<br> B. Sin40= x/6<br> C. Cos40= x/6<br> D. Cos40=6/x

STatiana [176]

The answer is A. sin40=6/x

7 0

3 years ago

Reduce to Standard form : (a) -21/91 (b) 32/(-256)

sweet [91]

Answer:

a) -3/13

b) -1/8

Step-by-step explanation:

a) - (21 / 7) / (91 / 7) = 3/13

b) (32 / 32 ) / - (256 / 32) = -1/8

5 0

2 years ago

For some medical imaging, the scale of the image is 5 : 1. That means that if an image is 5 cm long, the corresponding length on

svlad2 [7]

I believe it would be 1.5 cm squared because 7.5/5 = 5

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2 years ago

water level in a plastic pool changed by -8 gallons each hour after 6 hours,the pool contained 132 gallons.how much water was in

elena55 [62]

X = original amount of water
x - 8(6) = 132
x - 48 = 132
x = 132 + 48
x = 180...it originally contained 180 gallons

6 0

2 years ago

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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5 0

2 years ago

Read 2 more answers