(3 3/4) divided by (-2 1/2)

Suppose that 2 ≤ f '(x) ≤ 3 for all values of x. what are the minimum and maximum possible values of f(5) − f(1)?

zzz [600]

Assuming $f$ is differentiable over the interval (1, 5), by the mean value theorem we know that there is some $1$ such that

$f'(c)=\dfrac{f(5)-f(1)}{5-1}$

We're given that $2\le f'(x)\le3$ for all $x$ , so

$2\le\dfrac{f(5)-f(1)}4\le3\implies8\le f(5)-f(1)\le12$

that is, the minimum possible value is 8, and the maximum possible value is 12.

8 0

3 years ago

How are exponents and logarithms related? give an example with you explanation

Soloha48 [4]

Exponents and logarithms are inverse functions, which means that exponents are the way to "undo" logarithms and logarithms are what you use to "undo" exponents. exponents tell you how many times you have to multiply a value and logarithms ask which exponent you need to use to get a specified value. exponential equations are used in a lot of scientific fields; for example, in biology, they're used to calculate the half-life of organisms. they can also be used to estimate the growth of a population. additionally, the richter scale, which measures the magnitude of earthquakes, uses logarithms.

6 0

3 years ago

A supplier sells 214 2 1 4 pounds of mulch for every 113 1 1 3 pounds of gravel. The supplier sells 172 pounds of mulch and grav

Dennis_Churaev [7]

Answer:

2.25 + 1 1/3=3.5833333 combined number of pounds of mulch and gravel sold.

2.25/3.583333 X 172=108 pounds of mulch that the supplier sells

(1 1/3) / 3.583333 X 172=64 pounds of gravel that the supplier sells.

Step-by-step explanation:

Let 'x' = pounds of gravel The ratio of mulch to gravel is given 2 1/4 / 1 1/3 Which equals 9/4 / 4/3 = 27/16

So the total gravel and much is equal to x + 27/16x = 172

Solving for 'x'

43/16 x = 172

x = 172 (16/43) = 64 pounds of gravel Mulch = 27/16x = 64 (27/16)= 108 pounds

3 0

3 years ago

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

Two of the 240 passengers are chosen at random. Find the probability that

hjlf

Step-by-step explanation:

there are in total 240 passengers.

out of these 240, there are 150+30=180 passengers that are in holiday.

and 240-180 = 60 passengers are not.

if we pick one passenger then the probability is 180/240 = 3/4 = 0.75 that he/she is on holiday.

remember : desired "events" over total "events".

i)

now we pick 2 passengers.

the probabilty for the first one to be on holiday is again

3/4 or 0.75.

if that event happens, then we have only 179 passengers out of now 239 to be on holiday.

and to pick one out of that pool to be on holiday is then

179/239 = 0.748953975...

and for both events to happen in one scenario we need to multiply both probabilities (it is an "and" relation, while an addition would be for an "exclusive or" relation).

the probabilty that we pick 2 passengers on holiday is

3/4 × 179/239 = 0.561715481... ≈ 0.5617

we cannot simply square the basic probability of 0.75 (0.75² = 0.5625), because that would mean we pick one passenger, then put him back into the crowd, and then pick a second time (with a chance to pick the same person again). like with rolling a die.

but that is not the scenario as I understand it. it is to pick a passenger, then keep that person singled out and pick a second passenger. hence the difference.

ii)

exactly one of the two is in holiday.

that means

either the first one is on holiday and the second one is not, or the the second one is and the first one is not.

now we model this logic statement in probabilty arithmetic.

please note that after the first pull we need to update the numbers for the remaining pool depending on the result of the first pull.

the total remaining is in both cases 239. but either the remaining people on holiday go down to 179 (and not in holiday stays 60), or the remaining people not on holiday go down to 59 (and on holiday stays 180).

so, the first one is on holiday, and the second one is not :

3/4 × 60/239 (remember : "and" relation)

= 3 × 15/239 = 45/239 = 0.188284519...

the first one is not on holiday, and the second one is :

60/240 × 180/239 = 1/4 × 180/239 = 45/239 =

= 0.188284519...

since there is no overlap of the potential events (there is no event that could be in both cases), this is an exclusive or relation, and we can add the probabilities.

so, the probability for exactly one of the picked passengers to be on holiday is

2×0.188284519... = 0.376569038... ≈ 0.3766

8 0

2 years ago