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3 years ago

What is the midpoint of the x-intercepts of f(x) = (x - 2)(x – 4)?

Mathematics

1 answer:

ehidna [41]3 years ago

3 0

Answer:

Step-by-step explanation:

because it is.

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Evaluate the function at the given numbers( correct to six decimal places). Use the results to guess the value of the limit or e

netineya [11]

Answer:

Value of the limit is 0.5.

Step-by-step explanation:

Given,

$F(x)=\frac{e^x-1-x}{x^2}$

When,

$x=1,F(1)=frac{e^1-1-1}{1}=e-2=0.718281$

$x=0.5, F(0.5)=\frac{e^0.5-1-0.5}{(0.5)^2}=0.594885$

$x=0.1, F(0.1)=\frac{e^0.1-1-0.1}{(0.1)^2}=0.517091$

$x=0.05, F(0.05)=\frac{e^0.05-1-0.05}{(0.05)^2}=0.508438$

$x=0.01, F(0.01)=\frac{e^0.01-1-0.01}{(0.01)^2}=0.501670 \hfill (1)$

Correct upto six decimal places.

Now,

$\lim_{x\to 0}F(x)=\lim_{x\to 0}\frac{e^x-1-x}{x^2}$ $(\frac{0}{0})$ form, applying L-Hospital rule that is differentiating numerator and denominator we get,

$\lim_{x\to 0}F(x)$

$=\lim_{x\to 0}\frac{e^x-1}{2x}$ $(\frac{0}{0})$ form.

$=\lim_{x\to 0}\frac{e^x}{2}=\frac{1}{2}=0.5\hfill (2)$

Limit exist and is 0.5. That is according to (1) we can see as the value of x lesser than 1 and tending to near 0, value of the function decreases respectively. And from (2) it shows ultimately it decreases and reach at 0.5, consider as limit point of F(x).

8 0

3 years ago

I need help with this one math question please!

den301095 [7]

Answer:

see explanation

Step-by-step explanation:

Assuming the fractions are being multiplied

Factorise the denominators of both fractions

x² - 3x - 10 = (x - 5)(x + 2)

x² + x - 12 = (x + 4)(x - 3)

The product can now be expressed as

$\frac{x-5}{(x-5)(x+2)}$ × $\frac{x+2}{(x+4)(x-3)}$

Cancel (x - 5) and (x + 2) on the numerators/ denominators, leaving

$\frac{1}{(x+4)(x-3)}$ = $\frac{1}{x^2+x-12}$

3 0

3 years ago

Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,

vladimir1956 [14]

Answer:

(a) $C(6,0) = 1$ , $C(6,1) = 6$ , $C(6,2) = 15$ , $C(6,3) = 20$ , $C(6,4) = 15$ , $C(6,5) = 6$ , $C(6,6) = 1$ .

(b) $C(7,0) = 1$ , $C(7,1) = 7$ , $C(7,2) = 21$ , $C(7,3) = 35$ , $C(7,4) = 35$ , $C(7,5) = 21$ , $C(7,6) = 7$ , $C(7,7)=1$ .

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$ .

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$ , for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

$C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

$C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

$C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$ .

(b) The explanations to the solutions is just the calculations.

$C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

$C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

$C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$