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2 years ago

Brianna is solving the equation -5 + 6(6x + 6) = 1 + 6x. Her steps are shown.

Mathematics

1 answer:

GuDViN [60]2 years ago

8 0

Answer:

Part A: Brianna made a error on step 2. Because it has to be + 31 not + 41 cause it is a -5+36

Part B: x= -1

Step-by-step explanation:

This is how it should have been done.

Step 1: -5+6(6x+6)= 1+6x

Step 2: -5+36x= 36=1+6x

Step 3: 36x+31= 1+6x

Step 4: 36x= 6x-30

Step 5: 30x= -30

Step 6: x= -1

You might be interested in

Find the 88 term of the arithmetic sequence 4,-1,-6

tresset_1 [31]

Answer:

an = a1 + d(n - 1)

a88 = 4 + (-5)(88 - 1)

a88 = 4 + (-5)(87)

a88 = 4 - 435

a88 = -431

So, the 88th term of the arithmetic sequence is -431.

8 0

2 years ago

You have 14 coins that are nickles and quarters that equals 2 dollars how do you solve using substitution?

sergejj [24]

Answer:

The question is incorrect but if half nickels and quarters exist then you could say the answer is 7.5 nickels and 6.5 quarters.

The problem:

14 coins that are either nickels are quarters

The total dollar amount of these coins is 2 dollars

Step-by-step explanation:

Let n be the number of nickels.

Let q be the number of quarters.

We are given that n+q=14 and that .05n+.25q=2.

I'm going to choose to solve the system by elimination.

Multiplying the first equation by .25 gives:

.25n+.25q=.25(14)

.25n+.25q=3.5

So let's line the equations up now:

.25n+.25q=3.5

.05n+.25q=2

---------------------If we subtract the equations, the variable q will be eliminated. Thus, we will able to solve for the variable n.

.20n+0q=1.5

.20n=1.5

Dividing both sides by .2 gives:

n=7.5

If n=7.5 and n+q=14, then we have 7.5+q=14 by substitution property.

We can subtract 7.5 on both sides giving us:

q=14-7.5=6.5

We cannot have 7.5 nickels and 6.5 quarters but this satisfies the given information.

Check:

7.5+6.5=(7+6)+(.5+.5)=13+1=14

7.5(.05)+6.5(.25)=2

Now, the requested way which is by substitution:

Let n be the number of nickels.

Let q be the number of quarters.

We are given that n+q=14 and that .05n+.25q=2.

We can solve the first equation for either n or q and then substitute that into the other equation allowing us to solve for the other variable.

Let's solve for q by subtraction n on both sides:

q=14-n

We are going to replace q in the second equation with (14-n) giving us:

.05n+.25(14-n)=2

Distribute:

.05n+3.5-.25n=2

Combine like terms:

-0.2n+3.5=2

Subtract 3.5 on both sides:

-0.2n=-1.5

Divide both sides by -0.2:

n=7.5

Since q=14-n and n=7.5, then q=14-7.5=6.5 .

We are already check the solution above.

8 0

2 years ago

City A had a population of 10000 in the year 1990. City A’s population grows at a constant rate of 3% per year. City B has a pop

Georgia [21]

Answer:

City A and city B will have equal population 25years after 1990

Step-by-step explanation:

Given

Let

$t \to$ years after 1990

$A_t \to$ population function of city A

$B_t \to$ population function of city B

$A_0 = 10000$ ---- initial population (1990)

$r_A =3\%$ --- rate

$B_{10} = \frac{1}{2} * A_{10}$ ----- t = 10 in 2000

$A_{20} = B_{20} * (1 + 20\%)$ ---- t = 20 in 2010

Required

When they will have the same population

Both functions follow exponential function.

So, we have:

$A_t = A_0 * (1 + r_A)^t$

$B_t = B_0 * (1 + r_B)^t$

Calculate the population of city A in 2000 (t = 10)

$A_t = A_0 * (1 + r_A)^t$

$A_{10} = 10000 * (1 + 3\%)^{10}$

$A_{10} = 10000 * (1 + 0.03)^{10}$

$A_{10} = 10000 * (1.03)^{10}$

$A_{10} = 13439.16$

Calculate the population of city A in 2010 (t = 20)

$A_t = A_0 * (1 + r_A)^t$

$A_{20} = 10000 * (1 + 3\%)^{20}$

$A_{20} = 10000 * (1 + 0.03)^{20}$

$A_{20} = 10000 * (1.03)^{20}$

$A_{20} = 18061.11$

From the question, we have:

$B_{10} = \frac{1}{2} * A_{10}$ and $A_{20} = B_{20} * (1 + 20\%)$

$B_{10} = \frac{1}{2} * A_{10}$

$B_{10} = \frac{1}{2} * 13439.16$

$B_{10} = 6719.58$

$A_{20} = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 0.20)$

$18061.11 = B_{20} * (1.20)$

Solve for B20

$B_{20} = \frac{18061.11}{1.20}$

$B_{20} = 15050.93$

$B_{10} = 6719.58$ and $B_{20} = 15050.93$ can be used to determine the function of city B

$B_t = B_0 * (1 + r_B)^t$

For: $B_{10} = 6719.58$

We have:

$B_{10} = B_0 * (1 + r_B)^{10}$

$B_0 * (1 + r_B)^{10} = 6719.58$

For: $B_{20} = 15050.93$

We have:

$B_{20} = B_0 * (1 + r_B)^{20}$

$B_0 * (1 + r_B)^{20} = 15050.93$

Divide $B_0 * (1 + r_B)^{20} = 15050.93$ by $B_0 * (1 + r_B)^{10} = 6719.58$

$\frac{B_0 * (1 + r_B)^{20}}{B_0 * (1 + r_B)^{10}} = \frac{15050.93}{6719.58}$

$\frac{(1 + r_B)^{20}}{(1 + r_B)^{10}} = 2.2399$

Apply law of indices

$(1 + r_B)^{20-10} = 2.2399$

$(1 + r_B)^{10} = 2.2399$ --- (1)

Take 10th root of both sides

$1 + r_B = \sqrt[10]{2.2399}$

$1 + r_B = 1.08$

Subtract 1 from both sides

$r_B = 0.08$

To calculate $B_0$ , we have:

$B_0 * (1 + r_B)^{10} = 6719.58$

Recall that: $(1 + r_B)^{10} = 2.2399$

So:

$B_0 * 2.2399 = 6719.58$

$B_0 = \frac{6719.58}{2.2399}$

$B_0 = 3000$

Hence:

$B_t = B_0 * (1 + r_B)^t$

$B_t = 3000 * (1 + 0.08)^t$

$B_t = 3000 * (1.08)^t$

The question requires that we solve for t when:

$A_t = B_t$

Where:

$A_t = A_0 * (1 + r_A)^t$

$A_t = 10000 * (1 + 3\%)^t$

$A_t = 10000 * (1 + 0.03)^t$

$A_t = 10000 * (1.03)^t$

and

$B_t = 3000 * (1.08)^t$

$A_t = B_t$ becomes

$10000 * (1.03)^t = 3000 * (1.08)^t$

Divide both sides by 10000

$(1.03)^t = 0.3 * (1.08)^t$

Divide both sides by $(1.08)^t$

$(\frac{1.03}{1.08})^t = 0.3$

$(0.9537)^t = 0.3$

Take natural logarithm of both sides

$\ln(0.9537)^t = \ln(0.3)$

Rewrite as:

$t\cdot\ln(0.9537) = \ln(0.3)$

Solve for t

$t = \frac{\ln(0.3)}{ln(0.9537)}$

$t = 25.397$

Approximate

$t = 25$

7 0

3 years ago

A box of ice cream bars had a regular price of $60. The discount is 10% off the price. Calculate the new price after the discoun

marysya [2.9K]

60 * 0.10 = 6
$6 price off
60 - 6 = $54
Solution: new price: $54

6 0

2 years ago

Let A be the area of a circle with radius r. If dr/dt=3, find dA/dt when r=2.

astraxan [27]

$\bf \textit{area of a circle}=A=\pi r^2
\\\\\\
\cfrac{dA}{dt}=\pi \cdot 2r^1\cdot \cfrac{dr}{dt}\implies \cfrac{dA}{dt}=\pi \cdot 2r\cdot \cfrac{dr}{dt}\quad 
\begin{cases}
r=2\\
\frac{dr}{dt}=3
\end{cases}
\\\\\\
\cfrac{dA}{dt}=\pi \cdot 2(2)(3)$

3 0

3 years ago

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