Bonsoir pour vais vous m'aider svp développer et réduire l'expression D=(2x+3)²-(x+4)² et factoriser D Svp aider moi

Will someone explain what i did wrong here?

tensa zangetsu [6.8K]

The actual inverse function is:

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

And the domain is [0, ∞).

<h3>Where is the mistake?</h3>

Remember that for a given function f(x) with a domain D and a range R.

For the inverse function, f⁻¹(x) the domain is R and the range is D.

Here, for the given function the domain is x ≥ 3 and the range is [0, ∞).

Then for the inverse function, which is:

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

(to check this, you must have that):

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

The domain will be [0, ∞) and the range x ≥ 3

If you want to learn more about inverse functions:

brainly.com/question/14391067

#SPJ1

7 0

2 years ago

Tutorial Exercise A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due

ANTONII [103]

Answer:

Both the boats will closet together at 2:21:36 pm.

Step-by-step explanation:

Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).

Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),

Formula : d=v*t

at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)

the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)

Formula : D= $\sqrt{(x2-x1)^2+(y2-y1)^2}$

⇒ $D = \sqrt{20^2t^2+15(t-1)^2}$

Now let $F(t) = D^2(t)$

∵ $F'(t) = 800t + 450(t-1) = 1250t -450\\F'(t) =0$

⇒ t= 450/1250

⇒ t= .36 hours

⇒ = 21 min 36 sec

Since F"(t)=0,

∴ This time gives us a minimum.

Thus, The two boats will closet together at 2:21:36 pm.

3 0

3 years ago

A university found that of its students withdraw without completing the introductory statistics course. Assume that students reg

polet [3.4K]

Answer:

A university found that 30% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course.

a. Compute the probability that 2 or fewer will withdraw (to 4 decimals).

= 0.0355

b. Compute the probability that exactly 4 will withdraw (to 4 decimals).

= 0.1304

c. Compute the probability that more than 3 will withdraw (to 4 decimals).

= 0.8929

d. Compute the expected number of withdrawals.

= 6

Step-by-step explanation:

This is a binomial problem and the formula for binomial is:

$P(X = x) = nCx p^{x} q^{n - x}$

a) Compute the probability that 2 or fewer will withdraw

First we need to determine, given 2 students from the 20. Which is the probability of those 2 to withdraw and all others to complete the course. This is given by:

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 2) = 20C2(0.3)^2(0.7)^{18}\\P(X = 2) =190 * 0.09 * 0.001628413597\\P(X = 2) = 0.027845872524$

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 1) = 20C1(0.3)^1(0.7)^{19}\\P(X = 1) =20 * 0.3 * 0.001139889518\\P(X = 1) = 0.006839337111$

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 0) = 20C0(0.3)^0(0.7)^{20}\\P(X = 0) =1 * 1 * 0.000797922662\\P(X = 0) = 0.000797922662$

Finally, the probability that 2 or fewer students will withdraw is

$P(X = 2) + P(X = 1) + P(X = 0) \\= 0.027845872524 + 0.006839337111 + 0.000797922662\\= 0.035483132297\\= 0.0355$

b) Compute the probability that exactly 4 will withdraw.

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 4) = 20C4(0.3)^4(0.7)^{16}\\P(X = 4) = 4845 * 0.0081 * 0.003323293056\\P(X = 4) = 0.130420974373\\P(X = 4) = 0.1304$

c) Compute the probability that more than 3 will withdraw

First we will compute the probability that exactly 3 students withdraw, which is given by

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 3) = 20C3(0.3)^3(0.7)^{17}\\P(X = 3) = 1140 * 0.027 * 0.002326305139\\P(X = 3) = 0.071603672205\\P(X = 3) = 0.0716$

Then, using a) we have that the probability that 3 or fewer students withdraw is 0.0355+0.0716=0.1071. Therefore the probability that more than 3 will withdraw is 1 - 0.1071=0.8929

d) Compute the expected number of withdrawals.

E(X) = 3/10 * 20 = 6

Expected number of withdrawals is the 30% of 20 which is 6.

5 0

3 years ago

How do you read this expression 8(8x+7) and 8(3x)+8(7)

Misha Larkins [42]

Answer:

64x+56 & 24x+56

Step-by-step explanation:

I hope this is what you mean because this is not an equation as it is not set equal to anything.

Both problems solved by distributive property -