Sketch over the interval [0, 2π] the graph of: y = sin(x + π)

If 6!+7!+8!=(n)(6!), then the value of n is A) 15 B)16 C)63 D)64 E)15!

gladu [14]

6!+7!+8!=(n)(6!)
calculate the individual values first:
6!=720
7!=5040
8!=40320

plug them into the equation:
720+5040+40320=n720
solve for n
5760+40320=n720
46080=n720
divide both sides by 720 to isolate n
46080/720=n720/720
64=n

6 0

3 years ago

I’d appreciate the help!

Elis [28]

Answer:

$\displaystyle 1\frac{119}{250}\:liter$

$\displaystyle 7,5\:sleps$

$\displaystyle 37,3\:sleps$

Step-by-step explanation:

$\displaystyle 1\frac{119}{250} = \frac{1476}{1000}$

$\displaystyle 1\frac{1}{13} \times 7 = 7\frac{7}{13} ≈ 7,538461538 ≈ 7,5$

$\displaystyle \frac{41}{1\frac{1}{10}} = 37\frac{3}{11} ≈ 37,3$

I am joyous to assist you anytime.

4 0

3 years ago

John, Sally, and Natalie would all like to save some money. John decides that it

brilliants [131]

Answer:

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2) $y=100x+300$

Part 3) $\$12,300$

Part 4) $\$2,700$

Part 5) Is a exponential growth function

Part 6) $A=6,000(1.07)^{t}$

Part 7) $\$11,802.91$

Part 8) $\$6,869.40$

Part 9) Is a exponential growth function

Part 10) $A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

Part 11) $\$13,591.41$

Part 12) $\$6,107.01$

Part 13) Natalie has the most money after 10 years

Part 14) Sally has the most money after 2 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

y=mx+b

The slope is equal to

$m=\$100\ per\ month$

The y-intercept or initial value is

$b=\$300$

so

$y=100x+300$

therefore

John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

see part 1)

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

so

$10\ years=10(12)=120 months$

For x=120 months

substitute in the linear equation

$y=100(120)+300=\$12,300$

Part 4) How much money will John have after 2 years?

Remember that

1 year is equal to 12 months

so

$2\ years=2(12)=24\ months$

For x=24 months

substitute in the linear equation

$y=100(24)+300=\$2,700$

Part 5) What type of exponential model is Sally’s situation?

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$P=\$6,000\\ r=7\%=0.07\\n=1$

substitute in the formula above

$A=6,000(1+\frac{0.07}{1})^{1*t}\\ A=6,000(1.07)^{t}$

therefore

Is a exponential growth function

Part 6) Write the model equation for Sally’s situation

see the Part 5)

Part 7) How much money will Sally have after 10 years?

For t=10 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{10}=\$11,802.91$

Part 8) How much money will Sally have after 2 years?

For t=2 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{2}=\$6,869.40$

Part 9) What type of exponential model is Natalie’s situation?

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$P=\$5,000\\r=10\%=0.10$

substitute in the formula above

$A=5,000(e)^{0.10t}$

Applying property of exponents

$A=5,000(1.1052)^{t}$

therefore

Is a exponential growth function

Part 10) Write the model equation for Natalie’s situation

$A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

see Part 9)

Part 11) How much money will Natalie have after 10 years?

For t=10 years

substitute

$A=5,000(e)^{0.10*10}=\$13,591.41$

Part 12) How much money will Natalie have after 2 years?

For t=2 years

substitute

$A=5,000(e)^{0.10*2}=\$6,107.01$

Part 13) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

Part 14) Who will have the most money after 2 years?

Compare the final investment after 2 years of John, Sally, and Natalie

Sally has the most money after 2 years

3 0

4 years ago

Al sumar fracciones con igual denominador, debemos sumar numerador con numerador y denominador con denominador. Verdadero o fals

Ahat [919]

Answer:

La respuesta es falso.

Step-by-step explanation:

La respuesta es falso.

Cuando se suman fraccciones con igual denominador, se suman los numeradores (numerador con numerador) y se deja el mismo denominador (el cual es común en ambos). Por ejemplo, la suma de 1/5 + 3/5 da como resultado:

$\frac{1}{5} + \frac{3}{5} = \frac{1 + 3}{5} = \frac{4}{5}$

En el caso de fracciones con diferentes denominadores, tampoco se suma numerador con numerador y denominador con denominador. En ese caso se debe encontrar el mínimo común múltiplo.

Por lo tanto, la respuesta es falso.

Espero que te sea de utilidad!

7 0

3 years ago

In ΔABC, the measure of ∠C=90°, the measure of ∠B=64°, and BC= 6.6 Find the length of CA the nearest tenth of a foot

Marrrta [24]

Answer:

CA = 13.5

Step-by-step explanation:

tan 64 = CA/6.6

2.05 = CA/6.6

CA = 13.5

4 0

3 years ago

Sketch over the interval [0, 2π] the graph of: y = sin(x + π)​

Sketch over the interval [0, 2π] the graph of: y = sin(x + π)