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3 years ago

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Mathematics

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Mama L [17]3 years ago

4 0

Answer:

es no me interesa hací que nose

Step-by-step explanation:

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If F(x)=2x/7+4, which of the following of the inverse of F(x)?

lions [1.4K]

Hello,

Answer D

y=2x/7+4
x=2y/7+4
(x-4)*7/2=y

5 0

3 years ago

Read 2 more answers

A new car was valued at $28,000, and it's value depreciated to $14,000 over the next 3 years. Use the exponential equation for d

kompoz [17]

The exponential equation (in thousands) is:
$V = 28 e^{-rt}$
plug in t=3 yrs, and solve for 'r'.
$14 = 28 e^{-3r} \\ \\ \frac{1}{2} = e^{-3r} \\ \\ ln(\frac{1}{2}) = -3r \\ \\ r = \frac{ln(\frac{1}{2})}{-3} = 0.231 = 23.1%$

Find value of car when t = 10
$V = 28e^{-2.31} = 2.779$

8 0

3 years ago

An investment fund starts at $0 and grows at a rate of $100 per month. Another fund starts at $4000 and reduces by $720 per year

Pie

Answer: 2 years 1 month

Step-by-step explanation:

2 years 1 month at $100 a month = $2,500

$720 x 2 years = $1,440

$720 / 12 months = $60

$1,440 + $60 = $1,500

$4,000 - $1,500 = $2,500

6 0

3 years ago

Explain the relationship between the polynomial identities: x^2 − 1 = (x +1)(x − 1) and x^2 − a^2 =(x − a)(x +a).

Bess [88]

Answer:

So these two equation can be related by identity $x^2-a^2=(x+a)(x-a)$

Step-by-step explanation:

We have given equation $x^2-1=(x+1)(x-1)$

And $x^2-a^2=(x+a)(x-a)$

We know the algebraic identity $a^2-b^2=(a+b)(a-b)$

From this identity

$x^2-1$ can be written as (x+1) (x-1)

And using same identity $x^2-a^2$ can be written as $(x-a)(x-b)$

So these two equation can be related by identity $x^2-a^2=(x+a)(x-a)$

5 0

3 years ago

Find the linear approximating polynomial for the following function centered at the given point a.point a. b. Find the quadratic

ZanzabumX [31]

Answer:

$(a)L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\(b)Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\(c)L(-0.23\pi)=-0.6626\\Q(-0.23\pi)=-0.6613$

Step-by-step explanation:

Given the function:

$f(x)=sin x, a=-\frac{\pi}{4}$

(a)Linear approximating polynomial

$L(x)=f(a)+f'(a)(x-a)\\f(a)=sin(-\frac{\pi}{4})=-\frac{\sqrt{2} }{2}\\f'(x)=cos x, a=-\frac{\pi}{4}\\f'(a)=cos (-\frac{\pi}{4})=\frac{\sqrt{2} }{2} \\Therefore:\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x-(-\frac{\pi}{4}))\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})$

(b)Quadratic approximating polynomial

$Q(x)=L(x)+\frac{1}{2}f''(a)(x-a)^2\\f''(x)=-sin(x), \\f''(a)=-sin(-\frac{\pi}{4})=\frac{\sqrt{2} }{2}\\Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2$

(c)When $x=-0.23\pi$

Using Linear Approximation polynomial

$L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\L(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})\\L(-0.23\pi)=-0.6626$

Using the Quadratic approximating polynomial

$Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\Q(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(-0.23\pi+\frac{\pi}{4})^2=-0.6613$

4 0

3 years ago