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4 years ago

How do you solve an equation of 120=6x

Mathematics

2 answers:

yan [13]4 years ago

4 0

Answer:

x = 20

Step-by-step explanation:

Given

120 = 6x

Solve for x by dividing both sides by the multiplier 6

x = $\frac{120}{6}$ = 20

Ber [7]4 years ago

3 0

<h2>120 = 6x</h2><h2>÷6 ÷6 ← divide by 6 on both sides</h2><h2>20 = x</h2><h2 /><h2>Therefore, the value of x = 20.</h2>

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An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click

Sergeeva-Olga [200]

Answer:

A. $\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

B. $\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

Step-by-step explanation:

Given that:

An investment of Amount = $8000

earns at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A) Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

$A = Pe^{rt}$

where;

$A = (8000) \ e ^{0.7t}$

The instantaneous rate of change = $\dfrac{dA}{dt}$

$\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )$

$= 8000 \dfrac{d}{dt}e^{0.07 \ t}$

$\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}$

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 2 years; the instantaneous rate of change is:

$\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}$

$\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

$12000 = (8000)e^{0.07 \ t}$

$\dfrac{12000 }{8000}= e^{0.07 \ t}$

$1.5= e^{0.07 \ t}$

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 5.79

$\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}$

$\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

7 0

3 years ago

Will give brainliest!!! The MTC Theater holds 250 people, and last night’s performance sold out! Students under the age of 16 pa

Marina86 [1]

Answer:

Student's tickets: 60

Adult's tickets: 190

I think this is right

Step-by-step explanation:

"Let x represent the number of tickets sold to students under the age of 16. Let y represent those over the age of 16.

Equation 1, number of tickets sold:

The total number of tickets sold to students under the age of 16 plus the number of tickets sold to students over 16 must equal 250 (since the theater was sold out). The equation becomes:

x + y = 250

Equation 2, revenue from tickets sold.

The total revenue was $910. The cost of tickets under 16 is $2.50 and the cost of tickets over 16 is $4.00. The equation becomes:

2.50 * x + 4.00 * y = 910.00

The system of equations to solve is:

x + y = 250

2.50 * x + 4.00 * y = 910.00 "

5 0

3 years ago

An interior automotive supplier places several electrical wires in a harness.Apull test measures the force required to pull spli

oksano4ka [1.4K]

Answer:

a) For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assumin Normal distribution

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical dsitribution

The result is on the figure attached.

b) For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

Step-by-step explanation:

For this case we have the following data:

28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4

The quantile-quantile or q-q plot is a graphical procedure in order to check the validity of a distributional assumption for a data set. We just need to calculate "the theoretically expected value for each data point based on the distribution in question".

If the values are asusted to the assumed distribution, we will see that "the points on the q-q plot will fall approximately on a straight line"

For this case our distribution assumed is normal.

Part a

For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assuming Normal distribution (0,1)

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical distribution

The result is on the figure attached.

Part b

For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

4 0

4 years ago

Please help me <br>i neeed an help from you guys

Ymorist [56]

The ratio that isn’t equivalent to the others is 10/12 because 7/8 and 21/24 are equivalent to eachother

5 0

3 years ago

Read 2 more answers

Please help thanks!!

siniylev [52]

Givens

x = -2

y = 3

z = - 2

Expression

2x - 3y - z

Substitute and solve

2(-2) - 3(3) - (-2) Notice that the 3 has 1 minus and the 2 has 2 minus signs. That means the the -3*3 = - 9 and - -2 = 2

-4 - 9 + 2

- 13 + 2

-11

Answer: 1st Choice

4 0

4 years ago

How do you solve an equation of 120=6x​

How do you solve an equation of 120=6x