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3 years ago

What is the solution to this equation? 9^x – 1 = 2 A. 1 B. 2 C. -1/2 D. 1/2

Mathematics

2 answers:

choli [55]3 years ago

7 0

Answer:

Step-by-step explanation:

sukhopar [10]3 years ago

6 0

The answer is B for 9^x-1 =2

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I keep putting in the formula but I keep getting the answer wrong

Advocard [28]

Answer:

314 in² (nearest whole number)

Step-by-step explanation:

Radius of a regular polygon: The distance from the center of the polygon to any vertex. The radius of a hexagon is equal to the length of one side.

Therefore, from inspection of the given diagram:

radius = 11 in ⇒ side length = 11 in

To find the area of a regular polygon, we first need to calculate the apothem. The apothem is the line drawn from the center of the polygon to the midpoint of one of its sides.

$\textsf{Length of apothem (a)}=\dfrac{s}{2 \tan\left(\frac{180^{\circ}}{n}\right)}$

where:

s = length of one side
n = number of sides

Given:

s = 11 in
n = 6

Substitute the given values into the formula and solve for a:

$\implies \textsf{a}=\dfrac{11}{2 \tan\left(\frac{180^{\circ}}{6}\right)}=\dfrac{11\sqrt{3}}{2}$

Area of a Regular Polygon

$\textsf{A}=\dfrac{n\:s\:a}{2}$

where:

n = number of sides
s = length of one side
a = apothem

Given:

n = 6
s = 11

$\textsf{a}=\dfrac{11\sqrt{3}}{2}$

Substitute the given values into the formula and solve for A:

$\implies \sf A=\dfrac{6 \cdot 11 \cdot \dfrac{11\sqrt{3}}{2}}{2}$

$\implies \sf A=314.3672216...$

$\implies \sf A=314\:\:in^2\:\:(nearest\:whole\:number)$

8 0

2 years ago

Find f(2) if f(x) = (x + 1)2. 9 6 5 I'll give brainiest if right

solmaris [256]

Answer:

x=2

Step-by-step explanation:

6 0

3 years ago

Find the following:

Butoxors [25]

Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say $\displaystyle \lim_{x\rightarrow a}f(x)=L$ , if f(x) approaches L as x approaches 'a'.

(a)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\$

$\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

(b)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$