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max2010maxim [7]
3 years ago
13

a pizza shop offers 30% off the price of a large pizza every Tuesday night. If the regular price is $25, what is the discounted

price?"
Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0
Take the percentage and times it by the the price and you will get 30% OF the amount and all you have to do is subtract the 30% from your original price and you will get your answer which is $17.5

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The sum of the series 5(1/2)^i is x/1024 Then x =
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I just got this question. I think the answer is 5.
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GIVING BRAINLIEST AND 20 points!
mestny [16]

Answer:

The answer is 27

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

2(3y+6+3)=196−16

(2)(3y)+(2)(6)+(2)(3)=196+−16(Distribute)

6y+12+6=196+−16

(6y)+(12+6)=(196+−16)(Combine Like Terms)

6y+18=180

6y+18=180

Step 2: Subtract 18 from both sides.

6y+18−18=180−18

6y=162

Step 3: Divide both sides by 6.

6y

/6

=

162

/6

y=27

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2 years ago
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Leno4ka [110]

Answer:

3 units on the right and 2 on the left

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lozanna [386]

There are 0 x intercepts for this equation.

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3 years ago
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Mkey [24]

Answer:

X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

Step-by-step explanation:

The question we have at hand is, in other words,

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix} - where we have to solve for the value of X

If we have to isolate X here, then we would have to take the inverse of the following matrix ...

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix} ... so that it should be as follows ... \begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}

Therefore, we can conclude that the equation as to solve for " X " will be the following,

X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} - First find the 2 x 2 matrix inverse of the first portion,

\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1} = \frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}= \frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} ,

X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix} - Simplifying this, we should get ...

\begin{bmatrix}1&3\\ 2&4\end{bmatrix} ... which is our solution.

7 0
3 years ago
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