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geniusboy [140]
3 years ago
15

The temperature decreased overnight by 15.2°C to a temperature of –12.8°C. Suppose t represents the temperature before it decrea

sed.
Which equation can be used to represent this situation?

A.
t – 15.2 = –12.8

B.
t + 12.8 = –15.2

C.
t – 12.8 = 15.2

D.
t – (–15.2) = –12.8
Mathematics
1 answer:
VashaNatasha [74]3 years ago
8 0
A. initially or was that them it decreased by 15.2 and then it equaled -12.8
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lesya692 [45]

Answer:

1st option

Step-by-step explanation:

To find the difference of the given matrices, we just need to subtract the corresponding elements of the two matrices as shown below:

\left[\begin{array}{cc}-4&8\\3&12\end{array}\right] -\left[\begin{array}{cc}2&1\\-14&15\end{array}\right] \\\\ \\ =\left[\begin{array}{cc}-4-2&8-1\\3-(-14)&12-15\end{array}\right]\\\\ \\ =\left[\begin{array}{cc}-6&7\\17&-3\end{array}\right]

Thus, 1st option gives the correct answer

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4 years ago
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Wayne bought blueberries. He uses 3/8 of the blueberries to make blueberry bread, 1/6 of the blueberries to make pancakes, and 5
yKpoI14uk [10]

Answer:

Wayne bought 72 blueberries.

Step-by-step explanation:

To combine the fractions, convert all of the fractions to a denominator of 24.

3/8 * 3/3 = 9/24

1/6 * 4/4 = 4/24

5/12 * 2/2 = 10/24

Now add together the fractions

9/24 + 4/24 + 10/24 = 23/24

He used 23/24 of his blueberries, or 69 of them.

To find the total amount of blueberries, divide 69 by 23, then multiply that number b 24.

69 / 23 = 3

3 * 24 = 72

Wayne bought 72 blueberries.

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4 years ago
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A North American river otter wants to cross the North Saskatchewan river. It can swim at a speed of 2.75 m/s. In that location,
insens350 [35]

Answer:

Step-by-step explanation:

time to cross will be the distance divided by the velocity in that direction.

a) t = d/v = 45/2.75 = 16.363636... 16.4 s

In the same time that the otter is swimming cross stream, the stream is also moving perpendicularly at its own pace.

b) d = vt = 2.5(16.36) = 40.9090... 40.9 m

As velocity is a vector, An observer on the bank would see the velocity as the vector addition of the cross stream and down stream velocities

c) v = √(2.75² + 2.5²) = 3.716517... 3.72 m/s  magnitude

  θ = arctan2.5/2.75 = 42.2736... 42.3° downstream of a line straight across

7 0
3 years ago
Which algebraic expression is equivalent to the expression below? 4(5 + 8r) + 19
mr_godi [17]

Answer:

32r + 39

Step-by-step explanation:

<u>Step 1:  Distribute</u>

4(5 + 8r) + 19

<em>20 + 32r + 19</em>

<em />

<u>Step 2:  Combine like terms</u>

<em>32r + 39</em>

<em />

Answer:  32r + 39

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3 years ago
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A gold mine is projected to produce $52,000 during its first year of operation, $50,000 the second year, $48,000 the third year,
marshall27 [118]

Answer:

its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

Step-by-step explanation:

Given that;

gold mine is projected to produce $52,000 during its first year

produce $50,000 the second year

produce $48,000 the third year

the mine is expected to produce for 20yrs; i.e n = 20

annual interest rate = 4% = 0.04%

now let P represent the present worth

we determine the present worth ;

Present worth ⇒ Cashflow(Uniform series present worth) - (2000)(uniform gradient present worth)

⇒Cashflow(P | A,i%,n) - (2000)(P | G,i%,n)

= 52000[ ((1+i)ⁿ - 1) / (i(1+i)ⁿ) ] - (2000)[ {((1+i)ⁿ - 1) / (i²(1+i)ⁿ))} - (n/i(1 + i)ⁿ) ]

= 52000[ ((1+0.04)²⁰ - 1) / (0.04(1+0.04)²⁰) ] - (2000)[ {((1+0.04)20 - 1) / ((0.04)²(1+0.04)²⁰))} - (20/0.04(1 + 0.04)²⁰) ]

= 52000[ ((1.04)²⁰ - 1) / (0.04(1.04)²⁰) ] - (2000)[ {((1.04)²⁰ - 1) / ((0.04)²(1.04)²⁰))} - (20/0.04(1.04)²⁰) ]

= 52000[ ((2.191123 - 1) / (0.04(2.191123) ] - (2000)[ {((2.191123 - 1) / ((0.0016)(2.191123))} - (20/0.04(2.191123) ]

= 52000(1.191123/0.08764) - (2000){( 1.191123/0.003506) - (20/0.87645)}

= 52000(13.59033) - (2000)(339.7582 - 228.1935)

= 52000(13.59033) - (2000)(111.5647)

= 706695.6 - 223129.4

= 483,566.2 ≈ 483,566

Therefore its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

4 0
3 years ago
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