Whats the proper abbreviation for twenty-five micrograms

Which graph represents decreasing distance with increasing time? Distance Distance (my Distance (mi Distance (m) Time (min) Time

Leona [35]

Answer:

The second one

Step-by-step explanation:

5 0

4 years ago

Please answer this correctly

n200080 [17]

Answer:

Here you go

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

When rabbits were introduced to the continent of Australia they quickly multiplied and spread across the continent since there w

Lady bird [3.3K]

Answer:

a. $y=6(1.7472)^x$

b. $y=6e^{0.558t}$

c.13.3 months

Step-by-step explanation:

a.-Given the first term at $t_0$ is 6 and the second term at $t_3$ is 32.

-Let's take rabbit population as a function of time to be

$y=ab^x$

where y is the population at time x and a the initial population at $t_0\\$

#We substitute our values to calculate the value of the constant b:

$y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472$

#Replace b in the population function:

$y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x$

Hence, the regression for the rabbit population as a function of time x is $y=6(1.7472)^x$

b. The exponential function in terms of base $e$ is usually expressed as:

$A=A_0e^{kt}$

Where:

$A_0$ -is the initial population at $t_o$

$A$ -is the population at time t.

$k-$ is the exponential growth constant.

$e-$ the exponent

Our function in terms of base exponent is rewritten as:

$y=A_0e^{kt}$

#Substitute with actual figures to solve for t:

$y=A_0e^{kt}, y=32, xt=3, A_0=6\\\\32=6e^{3k}\\\\3k=In (32/6)\\\\k=0.5580$

Hence, the regression equation in terms of base e is $y=6e^{0.558t}$

c. We substitute y with any number higher than 10,000 to estimate the time for the rabbits to exceed 10,000.

-We know that $y=6e^{0.558t}.$

Therefore we calculate t as(take y=10001):

$y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951$

Hence, it takes approximately 13.3 months for the population to exceed 10000

3 0

4 years ago

Write an equation for the total water usage, U, at the park this year in terms of x, the number of special events the park has h

Hoochie [10]

Answer:

U = 16206 + 22.2x

Step-by-step explanation:

This year, water usage has increased by 20% for both regular days and special events.. Write an equation for the total water usage, U, at the park this year in terms of x, the number of special events the park has hosted

Solution:

Last year regular usage of water was 13505 gallons, where as it was 18.5 gallons per special event. Let us assume that there were x special invents, hence:

The total usage for last year = 13505 + 18.5x

This year, the total usage (U) increased by 20%, hence:

Total usage for this year (U) = total usage for last year + 20% of total usage for last year

U = (13505 + 18.5x) + 0.2(13505 + 18.5x)

U = 13505 + 18.5x + 2701 + 3.7x

U = 16206 + 22.2x

8 0

3 years ago

2.7·6.2–9.3·1.2+6.2·9.3–1.2·2.7 not pemdas

antiseptic1488 [7]

60

See steps

Step by Step Solution:

More Icon

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "2.7" was replaced by "(27/10)". 8 more similar replacement(s)

STEP

1

:

27

Simplify ——

10

Equation at the end of step

1

:

27 62 93 12 62 93 12 27

(((——•——)-(——•——))+(——•——))-(——•——)

10 10 10 10 10 10 10 10

STEP

2

:

6

Simplify —

5

Equation at the end of step

2

:

27 62 93 12 62 93 6 27

(((——•——)-(——•——))+(——•——))-(—•——)

10 10 10 10 10 10 5 10

STEP

3

:

93

Simplify ——

10

Equation at the end of step

3

:

27 62 93 12 62 93 81

(((——•——)-(——•——))+(——•——))-——

10 10 10 10 10 10 25

STEP

4

:

31

Simplify ——

5

Equation at the end of step

4

:

27 62 93 12 31 93 81

(((——•——)-(——•——))+(——•——))-——

10 10 10 10 5 10 25

STEP

5

:

6

Simplify —

5

Equation at the end of step

5

:

27 62 93 6 2883 81

(((——•——)-(——•—))+————)-——

10 10 10 5 50 25

STEP

6

:

93

Simplify ——

10

Equation at the end of step

6

:

27 62 93 6 2883 81

(((——•——)-(——•—))+————)-——

10 10 10 5 50 25

STEP

7

:

31

Simplify ——

5

Equation at the end of step

7

:

27 31 279 2883 81

(((—— • ——) - ———) + ————) - ——

10 5 25 50 25

STEP

8

:

27

Simplify ——

10

Equation at the end of step

8

:

27 31 279 2883 81

(((—— • ——) - ———) + ————) - ——

10 5 25 50 25

STEP

9

:

Calculating the Least Common Multiple

9.1 Find the Least Common Multiple

The left denominator is : 50

The right denominator is : 25

Number of times each prime factor

appears in the factorization of:

Prime

Factor Left

Denominator Right

Denominator L.C.M = Max

{Left,Right}

2 1 0 1

5 2 2 2

Product of all

Prime Factors 50 25 50

Least Common Multiple:

50

Calculating Multipliers :

9.2 Calculate multipliers for the two fractions

Denote the Least Common Multiple by L.C.M

Denote the Left Multiplier by Left_M

Denote the Right Multiplier by Right_M

Denote the Left Deniminator by L_Deno

Denote the Right Multiplier by R_Deno

Left_M = L.C.M / L_Deno = 1

Right_M = L.C.M / R_Deno = 2

Making Equivalent Fractions :

9.3 Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

L. Mult. • L. Num. 837

—————————————————— = ———

L.C.M 50

R. Mult. • R. Num. 279 • 2

—————————————————— = ———————

L.C.M 50

Adding fractions that have a common denominator :

9.4 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

837 - (279 • 2) 279

——————————————— = ———

50 50

Equation at the end of step

9

:

279 2883 81

(——— + ————) - ——

50 50 25

STEP

10

:

Adding fractions which have a common denominator

10.1 Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

279 + 2883 1581

—————————— = ————

50 25

Equation at the end of step

10

:

1581 81

———— - ——

25 25

STEP

11

:

Adding fractions which have a common denominator

11.1 Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

1581 - (81) 60

——————————— = ——

25 1

Final result :

60

7 0

3 years ago