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mars1129 [50]
3 years ago
11

Brad and his mom went out for lunch and the bill came to $25.45. His mom wanted to leave a 18% tip. How much tip should they lea

ve and what is the total bill
Mathematics
1 answer:
Oksana_A [137]3 years ago
3 0

Answer: they would leave a tip of

$4.581 and their total bill would be $30.031

Step-by-step explanation:

Brad and his mom went out for lunch and the bill came to $25.45. His mom wanted to leave a 18% tip. This means that the amount of the tip that she wanted to leave is 18% of the bill that came. Therefore

Amount of tip = 18/100 × the bill that came.

Amount of tip = 0.18×25.45= $4.581

So they would leave a tip of $4.581

behind.

The total bill would be sum of the original bill + sum of the amount of tip that they want to leave behind. Therefore, it becomes

25.45 + 4.581 = $30.031

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X 2 4 6 8 y 1 2 3 4  is a proportional relationship since it is going by multiples of 2 
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3 years ago
Please help
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Answer:

56°

Step-by-step explanation:

2x+3x+5=90

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5x=85

x=85÷5

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3 years ago
What is the equation of the line that passes through the point (8,4) and has a slope of 3/2
iren [92.7K]

Answer: Y= 3/2x-8=

Step-by-step explanation:Y=4 M= 3/2. X=-8

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4 = 3/2(8)+b

4= 24/2+b

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3 0
3 years ago
Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is
ohaa [14]

The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

The determinant is non-zero, so the vectors are indeed independent.

To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

p=ax^2+bx+c

Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

or the system (in matrix form)

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

Then

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}

\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}

A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

5 0
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anastassius [24]
.2 is 2/10 while .5 is 1/2.  1/2 is much larger than 1/5 (which is 2/10 reduced).
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