Use the function below to find f(-2)f(x)=5^x
A. 1/10
B. 1/25
C. -25
D. -10

Answers

Answer 1

Answer: $f(-2)=5^(-2)=(1)/(25)\Rightarrow \text{B}$

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Find the area of parallelogram ABCD given m2A=300 and the following measur 300 AX= 3 ft.; AB= 4V5 ft. A=

Answers

ABCD area =

AB = 4√2

AX= 3 ft = Height

then Area is

AREA= AB• AD • Sin 30°

THEN ANSWER IS

AREA= 4√2• 3 = 12√2 ft2

. = 17 square foots

For the following exercises, use this scenario: A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes. Rounding to six significant digits, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double

Answers

Answer:

$P(t)=256*e^(0.068185t)$

10 minutes for the population to double.

Step-by-step explanation:

The general formula for the population equation, P(t), is:

$P(t)=A*e^(nt)$

We are given that for t =5 minutes and t = 2 minutes:

$P(5)=360=A*e^(5n)\nP(20)=1,000=A*e^(20n)\n\n$

Solving for A and n:

$ln(360)=ln(A)+5n\nln(1,000)=ln(A)+20n\nln(1,000)-4ln(360)=ln(A)-4ln(A)+20n-20n\nln(A)=5.5455536\nA=256\nn=(ln(360)-ln(256))/(5)\nn=0.068185$

The exponential equation that represents this situation is:

$P(t)=256*e^(0.068185t)$

The population will double when P(t) = 512 bacteria:

$512=256*e^(0.068185t)\nln(2)=0.068185t\nt=10.17\ minutes$

To the nearest minute, it takes roughly 10 minutes for the population to double.

A construction worker tosses a brick from a tall building. The brick's height (in meters above the ground) t tt seconds after being thrown is modeled by h ( t ) = − 5 t 2 + 20 t + 105 h(t)=−5t 2 +20t+105h, left parenthesis, t, right parenthesis, equals, minus, 5, t, squared, plus, 20, t, plus, 105 Suppose we want to know the height of the brick above the ground at its highest point. 1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. h ( t ) = h(t)=h, left parenthesis, t, right parenthesis, equals 2) At its highest point, how far above the ground was the brick?

Answers

The vertex form of the equation is h(t) = -5 ( t - 2)² + 104.

At its highest point, the brick was 101 meters above the ground.

What is the vertex form of a quadratic function?

The factored form of a quadratic function is given as:

f(x) = a(x – h)2 + k

where a, h, and k are constants.

We have,

h(t) = -5t² + 20t + 105

h(t) = -5 ( t² - 4t ) + 105

h(t) = -5 ( t² - 4t + 2² ) - 2² + 105

h(t) = -5 ( t - 2 )² - 4 + 105

h(t) = -5 ( t - 2 )² + 101 _____(1)

The vertex form of the equation is h(t) = -5 ( t - 2)² + 104.

From equation (1) we can say that at its highest point the brick was at

101 meters above the ground.

Thus,

The vertex form of the equation is h(t) = -5 ( t - 2)² + 104.

At its highest point, the brick was 101 meters above the ground.

Learn more about quadraticequations here:

brainly.com/question/18804178

#SPJ5

Answer:

h(t)=−5(t−2)^2+125, 125

Step-by-step explanation:

Solve the equation. X^2-6x+6=0

Answers

$x^2-6x+6=0 \n \na=1, \ \ b=-6, \ \ c=6 \n\n\Delta =b^2-4ac = (-6)^2 -4\cdot1\cdot 6 = 36-24=12\n\n√(\Delta )=√(12)=√(4\cdot 3)=2√(3)\n \nx_(1)=(-b-√(\Delta) )/(2a)=(6-2√(3))/(2 )=(2( 3- √(3)))/(2)= 3- √(3)\n\nx_(2)=(-b+√(\Delta) )/(2a)=(6+2√(3))/(2 )=(2( 3+ √(3)))/(2)= 3+ √(3)$

$x^2-6x+6=0\n x^2-6x+9-3=0\n (x-3)^2=3\n x-3=\sqrt 3 \vee x-3=-\sqrt3\n x=3+\sqrt3 \vee x=3-\sqrt3$

Evaluate the integral: \[\int\limits_{}^{}ye ^{0.2y}dy\]
The problem we are having is the antiderivative of
\[e ^{0.2y}.\] We decided to split the problem into:
u = y
\[uv = e^{0.2y}\]
We just aren't sure where to go from here. Help?

Answers

$\int\limits_{}^{}ye ^(0.2y)dy\n\nu=y,du=dy\ndv=e ^{(y)/(5)}dy,v=\int\limits_{}^{}e ^{ (y)/(5) }dy=5e ^{ (y)/(5) }\n\n \int\limits_{}^{}ye ^{(y)/(5) }}dy=uv- \int\limits {v} \, du=y 5e ^{ (y)/(5) }-\int\limits {5e ^{ (y)/(5) }} \, dy= 5ye ^{ (y)/(5) }-25e ^{ (y)/(5) }=5e ^{ (y)/(5) }(y-5)$

Robert stands atop a 1,380-foot hill. He climbs down, reaching the bottom of the hill in 30 minutes. What is Robert’s average rate of elevation change in feet per minute?+460
+46
-46
-460

Answers

Before we calculate we can use some common sence thinknig to narrow down the choices. We know that Robert is gonig DOWN the hill, so it doesnt make sence that he woudl have a positive rate of change (i.e. the number feet up the hill he is is decreasing, not increasing) So right away, A & B are clearly wrong.

If we look at the last two (C & D) we can see that if -460 were right after 10 minutes he would have walked down 4,600 feet. This is WAY more that the total height of the hill and so can't be correct.

So C must be correct.

We can check this with some simple math:
$rate = (end-start)/(time) = (0ft-1380ft)/(30mins)= -46(ft)/(min)$