The function g(x) = 2^x. the function f(x) = 2^x+k and k< 0. which of the following statements is true.

2. There are 1500 candies in a candy dish. Each day half of the candy is taken out.

katen-ka-za [31]

Answer:

so easy and obvious dude

Step-by-step explanation:

5 0

3 years ago

Find the perimeter and area of each figure.

Sergio039 [100]

Answer:

1. A = 2x; P = 4x+2. A = 4; P = 10.

2. A = y² +2; P = 4y +2. A = 27; P = 22.

Step-by-step explanation:

1. The area is the sum of the marked areas of each of the tiles:

A = x + x

A = 2x

__

The perimeter is the sum of the outside edge dimensions of the tiles. Working clockwise from the upper left corner, the sum of exposed edge lengths is ...

P = 1 + (x-1) + x + 1 + (x+1) + x

P = 4x +2

__

When x=2, these values become ...

A = 2·2 = 4 . . . . square units

P = 4·2+2 = 10 . . . . units

_____

2. Again, the area is the sum of the marked areas:

A = y² + 1 + 1

A = y² +2

__

The edge dimension of the square y² tile is presumed to be y, so the perimeter (starting from upper left) is ...

P = y +(y-2) +1 +2 +(y+1) +y

P = 4y +2

__

When y=5, these values become ...

A = 5² +2 = 27 . . . . square units

P = 4·5 +2 = 22 . . . . units

4 0

3 years ago

2.

abruzzese [7]

Answer:

a) 30 kangaroos in 2030

b) decreasing 8% per year

c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo

Step-by-step explanation:

We assume your equation is supposed to be ...

P(t) = 76(0.92^t)

__

a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30

In the year 2030, the population of kangaroos in the province is modeled to be 30.

__

b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.

The population is decreasing by 8% each year.

__

c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.

P(100) = 75(0.92^100) = 76(0.0002392)

P(100) ≈ 0.0182, about 1/55th of a kangaroo

5 0

3 years ago

Consider an election with 681 votes a) If there are 5 candidates, what is the smallest number of first-place votes a candidate c

joja [24]

Answer:

137 votes

Step-by-step explanation:

considering an election with 681 votes and 5 candidates up for the election

dividing the votes among'st 5 candidates

= 681 / 5 = 136.2 hence the least number of first-place votes needed by a candidate using the plurality method would be = 137 votes

136.2 + 136.2 + 136.2 + 136.2 + 136.2 = 681 ( dividing the votes equally )

136 + 136 + 136 + 136+136 = 680

hence the remaining vote = 681 - 680 = 1

least first-place vote = 136 + 1 = 137 votes

6 0

3 years ago

Which expression is equivalent to (4^5/4*4^1/4 divided by 4^1/2) ^1/2

hoa [83]

$\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\\\ ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \left( \cfrac{4^{\frac{5}{4}}\cdot 4^{\frac{1}{4}}}{4^{\frac{1}{2}}} \right)^{\frac{1}{2}}\implies \left( \cfrac{4^{\frac{5}{4}\cdot \frac{1}{2}}\cdot 4^{\frac{1}{4}\cdot \frac{1}{2}}}{4^{\frac{1}{2}\cdot \frac{1}{2}}} \right)\implies \cfrac{4^{\frac{5}{8}}\cdot 4^{\frac{1}{8}}}{4^{\frac{1}{4}}}\implies \cfrac{4^{\frac{5}{8}+\frac{1}{8}}}{4^{\frac{1}{4}}}$

$\bf \cfrac{4^{\frac{6}{8}}}{4^{\frac{1}{4}}}\implies \cfrac{4^{\frac{3}{4}}}{4^{\frac{1}{4}}}\implies 4^{\frac{3}{4}}\cdot 4^{-\frac{1}{4}}\implies 4^{\frac{3}{4}-\frac{1}{4}}\implies 4^{\frac{2}{4}}\implies 4^{\frac{1}{2}}\implies \sqrt{4}\implies 2$

4 0

3 years ago