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3 years ago

What is the 13 term of the geometric sequence whit this explicit formula an=3•(-2)(n-1)

Mathematics

1 answer:

Orlov [11]3 years ago

3 0

Answer:

12,288

Step-by-step explanation:

aₙ = 3 (-2)ⁿ⁻¹

a₁₃ = 3 (-2)¹³⁻¹

a₁₃ = 12,288

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The function g is defined as follows for the domain given. g(x) = 2 - 2x , domain = \{- 1, 0, 1, 2\} Write the range of g using

Musya8 [376]

Explanation:

The function is given below as

$g(x)=2-2x$

The set of domains is given below as

$\lbrace-1,0,1,2\rbrace$

Step 1:

Put x=-1

$\begin{gathered} g(x)=2-2x \\ g(-1)=2-2(-1) \\ g(-1)=2+2 \\ g(-1)=4 \end{gathered}$

Step 2:

Put x=0

$\begin{gathered} g(x)=2-2x \\ g(0)=2-2(0) \\ g(0)=2 \end{gathered}$

Step 3:

Put x=1

$\begin{gathered} g(x)=2-2x \\ g(1)=2-2(1) \\ g(1)=2-2 \\ g(1)=0 \end{gathered}$

Step 4:

Put x=2

$\begin{gathered} g(x)=2-2x \\ g(2)=2-2(2) \\ g(2)=2-4 \\ g(2)=-2 \end{gathered}$

Hence,

The range of g(x) using set notation will be

$\Rightarrow\lbrace4,2,0,-2\rbrace$

Hence,

The graph of g(x) is given below as

8 0

1 year ago

Applying Properties of Exponents In Exercise,use the properties of exponents to simplify the expression.

vitfil [10]

Answer:

(A) $e^2$

(b) $e^{-3}$

(d) $e^3$

Step-by-step explanation:

We have given expression and we have to simplify the expression using exponent property

(A) $(\frac{1}{e})^{-2}$

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

(b) $(\frac{e^5}{e^2})^{-1}$

So $(\frac{e^5}{e^2})^{-1}=(e^{3})^{-1}=e^{-3}$

So $\frac{e^5}{e^3}=e^2$

(d) $\frac{1}{e^{-3}}=e^3$

7 0

3 years ago

In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm

tangare [24]

Answer:

$AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm$

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

$AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm$

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

$\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD$

Consider right triangle CAH.By the Pythagorean theorem,

$CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8$