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Agata [3.3K]
2 years ago
6

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP HELPELEPLEPE

Mathematics
2 answers:
diamong [38]2 years ago
7 0

Answer:

3/8

Step-by-step explanation:

sin A =  3/√73

cos A = √1 - ( 3/√73)²

          = √64/73

          = 8/√73    ... A in 1st quadrant cos A is positive

tan A = sin A/cos A = ( 3/√73) / ( 8/√73) = 3/8

ss7ja [257]2 years ago
3 0

Answer:

amogus

Step-by-step explanation:

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Find the solution to the following system using substitution or elimination: y = 3x + 2 y = -2x - 8 O A. (-5,-) OB. (-2,-4) O C.
mr Goodwill [35]

Answer:

B. (-2,-4)

Explanation

Given equations:

   y = 3x + 2

   y = -2x - 8

Solving both equations will yield the values of x and y;

Solution:

   y = 3x + 2    ----- (i)

   y = -2x - 8   ------ (ii)

Using substitution method, input equation i, into ii

    3x + 2 = -2x - 8

Collect like terms and solve;

     3x + 2x = -8 -2

         5x  = -10

            x  = -2

Then put x = -2 into i, to find y

      y = (-2 x 3) + 2

       y = -6 + 2 = -4

So, the solution of the equation is B. (-2,-4)

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3 years ago
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What is the domain of y= log_4(x+3)? all real numbers less than -3 all real numbers greater than –3 all real numbers less than 3
antiseptic1488 [7]

Step-by-step answer:

The domain of log functions (any legitimate base) requires that the argument evaluates to a positive real number.

For example, the domain of log(4x) will remain positive when x>0.

The domain of log_4(x+3) requires that x+3 >0, i.e. x>-3.

Finally, the domain of log_2(x-3) is such that x-3>0, or x>3.

7 0
3 years ago
A metallurgist has one alloy containing 32% copper and another containing 65% copper. How many pounds of each alloy must he use
kompoz [17]

Answer:

<h2>30.67 pounds of 32% copper alloy and 10.33 pounds of 65% copper alloy</h2>

Step-by-step explanation:

        First alloy contains 32% copper and the second alloy contains 65% alloy.

We wish to make 44 pounds of a third alloy containing 42% copper.

       Let the weight of first alloy used be x in pounds and the weight of second alloy used be y in pounds.

       Total weight = 44\text{ }pounds=x+y        -(i)

      Total weight of copper = 42\%\text{ of 44 pounds = }32\%\text{ of }x\text{ pounds + }65\%\text{ of }y\text{ pounds }

       \dfrac{42\times 44}{100}=\dfrac{32x}{100}+\dfrac{65y}{100}\\\\ 32x+65y=1848        -(ii)

       Subtracting 32 times first equation from second equation,

32x+65y-32x-32y=1848-32\times44\\33y=440\\y=13.333\text{ }pounds \\x=30.667\text{ }pounds

∴ 30.67 pounds of first alloy and 13.33 pounds of second alloy were used.

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3 years ago
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Answer:

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Step-by-step explanation:

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