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Katena32 [7]
2 years ago
11

5) A book is sold at $152.50 and a discount of 9% is given for students.

Mathematics
1 answer:
Svet_ta [14]2 years ago
4 0

Answer:

$138.78

Step-by-step explanation:

152.50 decrease 9% =

152.50 × (1 - 9%) = 152.50 × (1 - 0.09) = 138.775

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Solve Ax + By = -C for x.<br> Ox=-By-C<br> А<br> x=-By+C<br> Ox=By=0
dimulka [17.4K]

Answer:

\textbf{Option 1}\\

Step-by-step explanation:

\textup{Given $ Ax + By = -C$}\\\textup{Subtracting $By$ on both sides, we get,}\\$ Ax = -C -By $\\

\textup{Dividing by $A$ on both sides, (Assuming $A$ is not equal to zero) we get }\\$$ x = \frac{-By -C}{A}$$

3 0
2 years ago
Find the mode of the data set
SashulF [63]

Answer:

Rock-4

Country-6

Hip Hop-5

pop-3

Country is the mode

hope this helps

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Step-by-step explanation:

8 0
3 years ago
A bag contains 4 red marbles, 3 blue marbles and 3 yellow marbles. Match each statement with the correct value.
tia_tia [17]
I’m having a hard time understanding your question
4 0
3 years ago
Read 2 more answers
Help please! Picture included about functions. Explain if you can.
slega [8]
The definition of a function is a relation where every input (x value) is paired with exactly one output (y-value). 

A. Not a function, because you have (300, 9) and (300,7). The x-value 300 is paired with two different y-values, 9 and 7. 

B. Not a function. (260, 4) and (260, 2)

C. Is a function

D. Not a function. (225, 4) and (225, 2)
3 0
2 years ago
(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

4 0
3 years ago
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