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3 years ago

6 to the 5 power times 6 to the negative 2 power= ?

Mathematics

1 answer:

guapka [62]3 years ago

8 0

The answer the the question Is 5

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Multi step equations

miv72 [106K]

-86 = -6(-3p - 3) - 5p

Distribute the -6 into the parentheses:

-86 = (-6 * -3p) + (-6 * -3) - 5p

-86 = 18p + 18 - 5p

Simplify:

-86 = 13p + 18

Subtract both sides by 18:

-86 - 18 = 13p + 18 - 18

-104 = 13p

Divide both sides by 13:

$\frac{13p}{13}$ = $\frac{-104}{13}$

$\frac{p}{1}$ = $\frac{-8}{1}$

<u></u>

Good luck! If you have any questions, don't be afraid to ask :))

-T.B.

6 0

3 years ago

Read 2 more answers

Explain how to multiply the following whole numbers 21 x 14

Lesechka [4]

Answer:

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Step-by-step explanation:

Given

$21\:\times \:14$

Line up the numbers

$\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}$

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

Multiply the bold numbers: 1×4=4

$\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}$

Multiply the bold numbers: 2×4=8

$\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the top number by the bolded digit of the bottom number

$\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the bold numbers: 1×1=1

$\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}$

Multiply the bold numbers: 2×1=2

$\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}$

Add the rows to get the answer. For simplicity, fill in trailing zeros.

$\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}$

adding portion

$\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}$

Add the digits of the right-most column: 4+0=4

$\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}$

Add the digits of the right-most column: 8+1=9

$\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}$

Add the digits of the right-most column: 0+2=2

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Therefore,

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

6 0

3 years ago

If a = and b = -1 what is the value of ab -b^2

svet-max [94.6K]

Answer:

-2

Step-by-step explanation:

To solve you need to replace the variables with their numbers.

3(-1) - (-1^2)

-1 x -1 = -1

3 x (-1) = -3

-3 - (-1) =

-2

8 0

3 years ago

Trigonometry

saul85 [17]

Answer:

9.7

Step-by-step explanation:

tan=opposite/adjacent

tan=9/adjacent

tan 43=9/x

x tan 43=9

0.9325x=9

0.9325x/0.9325=9/0.9325

=9.65

hence;

x=9.7

7 0

3 years ago

Before the pandemic cancelled sports, a baseball team played home games in a stadium that holds up to 50,000 spectators. When ti

spayn [35]

Answer:

(a) $D(x)=-2,500x+60,000$

(b) $R(x)=60,000x-2500x^2$

(d)Optimal ticket price: $12

Maximum Revenue:$360,000

Step-by-step explanation:

The stadium holds up to 50,000 spectators.

When ticket prices were set at $12, the average attendance was 30,000.

When the ticket prices were on sale for $10, the average attendance was 35,000.

(a)The number of people that will buy tickets when they are priced at x dollars per ticket = D(x)

Since D(x) is a linear function of the form y=mx+b, we first find the slope using the points (12,30000) and (10,35000).

$\text{Slope, m}=\dfrac{30000-35000}{12-10}=-2500$

Therefore, we have:

$y=-2500x+b$

At point (12,30000)

$30000=-2500(12)+b\\b=30000+30000\\b=60000$

Therefore:

$D(x)=-2,500x+60,000$

(b)Revenue

$R(x)=x \cdot D(x) \implies R(x)=x(-2,500x+60,000)\\\\R(x)=60,000x-2500x^2$

(c)To find the critical values for R(x), we take the derivative and solve by setting it equal to zero.

$R(x)=60,000x-2500x^2\\R'(x)=60,000-5,000x\\60,000-5,000x=0\\60,000=5,000x\\x=12$

The critical value of R(x) is x=12.

(d)If the possible range of ticket prices (in dollars) is given by the interval [1,24]

Using the closed interval method, we evaluate R(x) at x=1, 12 and 24.

$R(x)=60,000x-2500x^2\\R(1)=60,000(1)-2500(1)^2=\$57,500\\R(12)=60,000(12)-2500(12)^2=\$360,000\\R(24)=60,000(24)-2500(24)^2=\$0$

Therefore:

Optimal ticket price:$12
Maximum Revenue:$360,000

3 0

3 years ago