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valentina_108 [34]
3 years ago
5

A computer store buys a computer system at a cost of 474.60 ​$. The selling price was first at 791 ​, but then the store adverti

sed a markdown on the system. Answer parts a and b. THIS IS ALSO DUE TONIGHT!!
Mathematics
1 answer:
GalinKa [24]3 years ago
5 0
So the question is the original price was $791 and they bought it at the marked down price at $474.60? then they saved $316.40 but i’m not really sure if that’s the question lol
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Find the GCF of 35, 60, and 25
user100 [1]

Answer:

<u>5</u> is the GCF

Step-by-step explanation:

The factors of 35 are: 1,5,7,35

The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

The factors of 25 are:  1, 5, 25

The highest or greatest factors that all of them have in common is 5.

4 0
3 years ago
What is the simplified form of this expression?<br> (7 + x) + (487-4-1)
tamaranim1 [39]

Answer:

Step-by-step explanation:

( 7 + ) + ( 4 8 7 − 4 − 1 ) (x+ 7 ) + (4 8 7− 4 − 1 )

( + 7 ) + 4 8 2 x+ 7 + 4 8 2

Add numbers

+ 48 9

Answer is x+489

3 0
3 years ago
3, -2, -7,...<br> what is the common difference of this arithmetic sequence
worty [1.4K]

Answer: -5

Step-by-step explanation: The common difference is the same number we are adding each time to get from one term to the next.

In this sequence, we are adding -5 each time

so our common difference is -5.

7 0
3 years ago
There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of
Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta (1)

Where:

AC - The distance from A to C, measured in miles.

AB - The distance from A to B, measured in miles.

BC - The distance from B to C, measured in miles.

\theta - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta

\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}

\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right) (2)

If we know that BC = 7\,mi, AB = 8\,mi, AC = 5\,mi, then the bearing from island B to island C:

\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]

\theta \approx 38.213^{\circ}

The bearing needed to navigate from island B to island C is approximately 38.213º.

8 0
3 years ago
Pls help me figure this out
hodyreva [135]

Answer:

C. \frac{27}{225}=\frac{b}{100}

Step-by-step explanation:

To find the number of black ribbons one would have subtract the amount of red ribbons from the total number of ribbons. Then one would have to express it as a proportion. Part over whole equals percent over 100. The number of black ribbons over total ribbons equals "b" over 100.

8 0
3 years ago
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