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vesna_86 [32]
2 years ago
8

Alexandra buys sweatshirts for $12 each. In her store, she sells each sweatshirt for $30.

Mathematics
2 answers:
serious [3.7K]2 years ago
7 0
Multiply $30 by .25 you get, it's $7.50so you subtract $7.50 from $30 and you get $22.50. Then multiply 2 (sweaters) by $22.50 and you get $45,but wait you still have to multiply that sales tax so you multiply .04 by $45 and you get $1.80, later you subtract $1.80 from $45 and you get $43.20(the cost of the two sweaters) I did part1 only cuz' no I have no time left Sorry for the inconvenience
Anit [1.1K]2 years ago
7 0

Answer:

Isn't 46.80????

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Ria can paint a room in 4 hours. Destiny can paint the same room in 6 hours. How long would it take Ria and Destiny to paint the
Lerok [7]

Time taken by Ria to paint a room = 4 hours

Time taken by Destiny to paint a room = 6 hours

If they work together, they complete 1 job. So,

\frac{1x}{4}+\frac{1x}{6}=1

The LCD of 4 and 6 is 12

\frac{3x+2x}{12}=1

\frac{5x}{12}=1

5x=12

x=\frac{12}{5}

Hence, both of them can paint the room in \frac{12}{5} hours or 2.4 hours.


4 0
2 years ago
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The product of three integers is −5. Determine all of the possible values for the three factors.
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-1,1,5
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6 0
2 years ago
Let P2 be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by 10x2+4xâ1, 3xâ4x2+3, and
lord [1]

I suppose

H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}

The vectors that span H form a basis for P_2 if they are (1) linearly independent and (2) any vector in P_2 can be expressed as a linear combination of those vectors (i.e. they span P_2).

  • Independence:

Compute the Wronskian determinant:

\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of H is 3.

  • Span:

Write an arbitrary vector in P_2 as ax^2+bx+c. Then the given vectors span P_2 if there is always a choice of scalars k_1,k_2,k_3 such that

k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c

which is equivalent to the system

\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}

so the vectors do span P_2.

The vectors comprising H form a basis for it because they are linearly independent.

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2 years ago
Tom buys a torch and a battery.
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Answer:

battery costs: 0.39...............

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Answer:its 275 for 1, and 325 for 2

Step-by-step explanation:

i did it

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2 years ago
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