HELP PLEASE I WILL MARK YOU AS A BRAINLIEST THANK YOU!!!

Solve the equation tanxtan(x – 45°)=0 for 0° < x < 360°.

Lady_Fox [76]

Answer:

\tan \left(\frac{\sin \left(x\right)-\cos \left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}\right)

Step-by-step explanation:

4 0

3 years ago

8775 find the prime factorization of the number

OLEGan [10]

8775

/ \

3 2925

/ \

3 975

/ \

3 325

/ \

5 65

/ \

5 13

/ \

13 1

Prime Factors: 3, 3, 3, 5, 5, 13

5 0

3 years ago

What number would you add to the equation below to complete the square?x2 + 4x = 0?

julia-pushkina [17]

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Equation
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x² + 4x = 0

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Add a constant
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x² + 4x + (4/2)² - (4/2)² = 0 ← Add 2² to form perfect square

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Evaluate bracket
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x² + 4x + 2² - 2² = 0

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Form the perfect square
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(x + 2)² = 4

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Square root both sides
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(x+2) = $\pm$ √4

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Evaluate the square root
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(x+2) = $\pm$ 2

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Solve x
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x = 2 - 2 or x = -2 - 2

x = 0 or -4

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Answer: The constant to add is 4 and the value of x = 0 or -4
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7 0

3 years ago

3/4x + 4/7y = 5/4x -1...solve for y

Zielflug [23.3K]

Answer:

$\frac{3}{4} x + \frac{4}{7} y = \frac{5}{4} x - 1 \\ \frac{4}{7} y = \frac{5}{4} x - 1 - \frac{3}{4} x \\ \\ \frac{4}{7} y = \frac{5x - 4 - 3x}{4} \\ \frac{4}{7} y = \frac{2x - 4}{4} \\ \\ y = \frac{2x - 4}{4} \div \frac{4}{7} \\ y = \frac{2x - 4}{4} \times \frac{7}{4} \\ \\ y = \frac{14x - 28}{16} \\ \\ y = \frac{14x}{16} - \frac{28}{16}$

I hope I helped you^_^

5 0

2 years ago

Input

melamori03 [73]

If we graph the given points then they appear to be in the shape of cubic function so we can use standard formula of cubic function which is

$y=ax^3+bx^2+cx+d$

Plug the given points to get four equations

like first point (3,6) so plug x=3 and y=6 into above equation, we get:

$6=a*3^3+b*3^2+c*3+d$

or

6=27a+9b+3c+d

Same way by plugging other points form table, we get total four equations as follows:

6=27a+9b+3c+d, 11=125a+25b+5c+d, 11=64a+16b+4c+d, 21=343a+49b+7c+d

Now we just need to solve these equations for a, b, c and d.

Solving them to get values of a, b, c and d, we get:

$a=\frac{25}{24}, b=-15, c=\frac{1715}{24}, d=-\frac{203}{2}$

Now plug them into

$y=ax^3+bx^2+cx+d$

So the required equation will be:

$y=\frac{25}{24}x^3-15x^2+\frac{1715}{24}x-\frac{203}{2}$

Hence required equation in function notation can be written as

$f(x)=\frac{25}{24}x^3-15x^2+\frac{1715}{24}x-\frac{203}{2}$

Now to prove that above function is correct, we just graph the given points from table and the obtained function.

After graphing the given points from table and the obtained function, We see that points lie on the graph of $y=\frac{25}{24}x^3-15x^2+\frac{1715}{24}x-\frac{203}{2}$

Which proves that our equation is correct.

5 0

3 years ago