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Elan Coil [88]
3 years ago
5

Why do you think the coefficient is called constant of variation?

Mathematics
1 answer:
kicyunya [14]3 years ago
3 0
The coefficient is called a constant
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Simplify the expression.
MAVERICK [17]
The answer is 6t^2+2.5x^2-0.5t-5.4
5 0
3 years ago
What is the least common multiple<br> of 1/3 5/8 1/2?
joja [24]

Answer: The answer is 16.

Step-by-step explanation:

The least common denominator means the same as least common multiple.

3, 6, 9, 12, 16

8, 16

2, 4, 6, 8, 10, 12, 14, 16y-step explanation:

7 0
3 years ago
Which function has the same y-intercept as the graph below ?
Natalija [7]

Answer:

D. y+4=2x

Step-by-step explanation:

-4 from both sides

y=2x-4

        ^ y-intercept (-4) same as on the graph          

4 0
3 years ago
What is the solution to the equations represented by these two lines
salantis [7]

Look at the picture.

A.) (3, 1)

6 0
3 years ago
Find the values of sin2u, cos2u, and tan2u given the figure.
Vadim26 [7]

Givens

y = 2

x = 1

z(the hypotenuse) = √(2^2 + 1^2)  = √5

Cos(u) = x value / hypotenuse = 1/√5

Sin(u) = y value / hypotenuse = 2/√5

Solve for sin2u

Sin(2u) = 2*sin(u)*cos(u)

Sin(2u) = 2(\dfrac{1}{\dsqrt{5}} * \frac{2}{\dsqrt{5}} = \dfrac{2}{5}) = 4/5

Solve for cos(2u)

cos(2u) = - sqrt(1 - sin^2(2u))

Cos(2u) = - sqrt(1 - (4/5)^2 )

Cos(2u) = -sqrt(1 - 16/25)

cos(2u) = -sqrt(9/25)

cos(2u) = -3/5

Solve for Tan(2u)

tan(2u) = sin(2u) / cos(2u) = 4/5// - 3/5 = - 0.8/0.6 = - 1.3333 = - 4/3

Notes

One: Notice that you would normally rationalize the denominator, but you don't have to in this case.  The formulas are such that they perform the rationalizations themselves.

Two: Notice the sign on the cos(2u). The sin is plus even though the angle (2u) is in the second quadrant. The cos is different. It is about 126 degrees which would make it a negative root (9/25)

Three: If you are uncomfortable with the  tan, you could do fractions.

\text{tan(2u)} = \dfrac{\dfrac{4}{5}}{\dfrac{-3}{5} } =\dfrac{4}{5} *\dfrac{5}{-3} =\dfrac{4}{-3}

7 0
3 years ago
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