All categories

3 years ago

4. solve e) 4^x-10.2^x+16=0 I think the dot means multiply?

2 answers:

6 0

$4^x-10\cdot2^x+16=0\Rightarrow(2^2)^x-10\cdot2^x+16=0\Rightarrow(2^x)^2-10\cdot2^x+16=0$

$subtitute\ 2^x=t > 0\\\\t^2-10t+16=0\\t^2-2t-8t+16=0\\t(t-2)-8(t-2)=0\\(t-2)(t-8)=0\iff t-2=0\ or\ t-8=0\\\\t=2\ or\ t=8\\\\t=2^1\ or\ t=2^3\\\\therefore\ 2^x=2^1\ or\ 2^x=2^3\\\\Answer:\boxed{x=1\ or\ x=3}$

zubka84 [21]3 years ago

3 0

2^x=a
4^x=(2^2)^x=(2^x)^2=a^2

a^2-10a+16=0
Δ=10^2-4*1*16=100-64=36
√Δ=6
a1=(10+6)/2=16/2=8
a2=(10-6)/2=4/2=2
2^x=8=2^3⇒ x=3 or
2^x=2=2^1⇒ x=1

There are 2 solutions : x=1 or x=3

You might be interested in

Donut $2.50

anastassius [24]

Bro don’t click the link ☝️

4 0

2 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago

Find the equation of the line that is parallel and passes through the point (9,6) find the equation of the line perpendicular

masya89 [10]

Any line parallel to this line can be written as y=-2/3x+c and passes through (9,6).thus 6=-2/3(9)+c.c-6=6.c=12.equation is y=-2/3x+12.

3 0

2 years ago

CAN SOMEONE GOOD AT MATH HELP PLEEESE!!

Rufina [12.5K]

Answer: 1 .Thus for a graph to have an Euler circuit, all vertices must have even degree. The converse is also true: if all the vertices of a graph have even degree, then the graph has an Euler circuit, and if there are exactly two vertices with odd degree, the graph has an Euler path.

2. A graph has an Euler circuit if and only if the degree of every vertex is even. A graph has an Euler path if and only if there are at most two vertices with odd degree.

Step-by-step explanation:

Can i have branily plz

7 0

2 years ago

Tests for tuberculosis like all other diagnostic tests are not perfect. QFT-G is one of such tests for tuberculosis. Suppose tha

Sauron [17]

Answer:

the probability of an adult getting a NEG result and truly having tuberculosis is 0.0127

Step-by-step explanation:

S = the adult really has tuberculosis.

S' = complement of S = the adult does not has tuberculosis.

POS = the test gives a positive result

P(S)= 0.05

P(POS | S)=0.746

P(NEG | S')= 0.7653

this is an intersection because the "and" word

P(NEG ∩ S) = P(NEG| S)*P(S)=(1-P(POS | S))*P(S)=(1-0.746)*0.05=0.0127

6 0

2 years ago