151% find the decimal notation

I need help with number 4

ss7ja [257]

Answer: Choice D

$f^{-1}(x) = (x-3)^2+2\\\\$

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Explanation:

First we replace f(x) with y. This is because both y and f(x) are outputs of a function.

To find the inverse, we swap x and y and solve for y like so

$y = \sqrt{x-2}+3\\\\x = \sqrt{y-2}+3 \ \text{ .... swap x and y; isolate y}\\\\x-3 = \sqrt{y-2}\\\\(x-3)^2 = y-2 \ \text{ ... square both sides}\\\\(x-3)^2+2 = y\\\\y = (x-3)^2+2\\\\f^{-1}(x) = (x-3)^2+2\\\\$

Note: because the range of the original function is $y \ge 3$ , this means the domain of the inverse is $x \ge 3$ . The domain and range swap roles because of the swap of x and y.

As the graph shows below, the original and its inverse are symmetrical about the mirror line y = x. One curve is the mirror image of the other over this dashed line.

7 0

3 years ago

Read 2 more answers

Given: KLMN is a trapezoid, m∠N= m∠KML, FD=8, LM KN = 3/5 F∈ KL , D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3 root5 Find: KM

Alla [95]

Answer:

The length of KM is $\sqrt{109}$ units.

Step-by-step explanation:

Given information: KLMN is a trapezoid, ∠N= ∠KML, FD=8, $\frac{LM}{KN}=\frac{3}{5}$ , F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, $ME=3\sqrt{5}$ .

From the given information it is noticed that the point F and D are midpoints of KL and MN respectively. The height of the trapezoid is $3\sqrt{5}$ .

Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.

Since $\frac{LM}{KN}=\frac{3}{5}$ , therefore LM is 3x and KN is 5x.

$\frac{3x+5x}{2}=8$

$\frac{8x}{2}=8$

$x=2$

Therefore the length of LM is 6 and length of KN is 10.

Draw perpendicular on KN form L and M.

$KN=KA+AE+EN$

$10=6+2(EN)$ (KA=EN, isosceles trapezoid)

$EN=2$

$KE=KN-EN=10-2=8$

Therefore the length of KE is 8.

Use pythagoras theorem is triangle EKM.

$Hypotenuse^2=base^2+perpendicular^2$

$(KM)^2=(KE)^2+(ME)^2$

$(KM)^2=(8)^2+(3\sqrt{5})^2$

$KM^2=64+9(5)$

$KM=\sqrt{109}$

Therefore the length of KM is $\sqrt{109}$ units.

5 0

3 years ago

Exercise 2.2.5. Let [[x]] be the greatest integer less than or equal to x. For example, [[π]] = 3 and [[3]] = 3. For each sequen

Debora [2.8K]

Answer:

a) $\lim_{n\rightarrow \infty}a_n=0$

b) The limit is [[4/3]]=1, that is, $\lim_{n\rightarrow \infty}a_n=1$

Step-by-step explanation:

The definition of the limit of a sequence (an) is as follows:

$\lim_{n\rightarrow \infty}a_n=L$ (the sequence (an) converges to L) means that given $\epsilon>0$ , there exists some natural number N such that if n≥N, then $|a_n-L|$

We will use that [[x]] is a increasing function. That is, if x≤y then [[x]]≤[[y]]

a) For all $\epsilon>0$ , take N≥6. Then 1/N ≤1/6 and 5/N ≤ 5/6 < 1. We have that 0<5/N<1, therefore [[5/N]]=0 (0 is the biggest integer less than or equal than x, for any x on the interval 0<x<1)

For n≥N, we have that 1/n ≤ 1/N, hence 0< 5/n ≤ 5/N < 1 . As we discussed above, [[5/n]]=0. Then |a_n|=0, therefore $|a_n-0|=|a_n|=0$