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3 years ago

PLEASE HELO ASAP WILL MARK BRAINLIEST

Mathematics

1 answer:

ss7ja [257]3 years ago

6 0

A.) 1/2

B.) 3

For the first one, just plug the value of x (4) everywhere you see x in the formula.

For the second one, place 1 wherever you see f(x)

Hope this helps

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IF YOU DO IT THEN YOU WILL GET BRAINLIEST! <br> −9x+4y=6<br> 9x+5y=−33<br> <br> SOLVE FOR X AND Y

BigorU [14]

Answer:

x = 2 y = 3

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

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4 years ago

Does anyone know how to factor this?

NNADVOKAT [17]

3(4x-3)(x+6) is the answer

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3 years ago

Base: z(x)=cosx Period:180 Maximum:5 Minimum: -4 What are the transformation? Domain and Range? Graph?

garik1379 [7]

Answer:

The transformations needed to obtain the new function are horizontal scaling, vertical scaling and vertical translation. The resultant function is $z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)$ .

The domain of the function is all real numbers and its range is between -4 and 5.

The graph is enclosed below as attachment.

Step-by-step explanation:

Let be $z (x) = \cos x$ the base formula, where $x$ is measured in sexagesimal degrees. This expression must be transformed by using the following data:

$T = 180^{\circ}$ (Period)

$z_{min} = -4$ (Minimum)

$z_{max} = 5$ (Maximum)

The cosine function is a periodic bounded function that lies between -1 and 1, that is, twice the unit amplitude, and periodicity of $2\pi$ radians. In addition, the following considerations must be taken into account for transformations:

1) $x$ must be replaced by $\frac{2\pi\cdot x}{180^{\circ}}$ . (Horizontal scaling)

2) The cosine function must be multiplied by a new amplitude (Vertical scaling), which is:

$\Delta z = \frac{z_{max}-z_{min}}{2}$

$\Delta z = \frac{5+4}{2}$

$\Delta z = \frac{9}{2}$

3) Midpoint value must be changed from zero to the midpoint between new minimum and maximum. (Vertical translation)

$z_{m} = \frac{z_{min}+z_{max}}{2}$

$z_{m} = \frac{1}{2}$

The new function is:

$z'(x) = z_{m} + \Delta z\cdot \cos \left(\frac{2\pi\cdot x}{T} \right)$

Given that $z_{m} = \frac{1}{2}$ , $\Delta z = \frac{9}{2}$ and $T = 180^{\circ}$ , the outcome is:

$z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)$

The domain of the function is all real numbers and its range is between -4 and 5. The graph is enclosed below as attachment.

8 0

3 years ago

Multiple choice:

Paha777 [63]

In the given you wrote -2, whereas it should be -2i

z = a + bi and in trigo form | z |.(cos Ф + i.sin Ф).
z = 0 -2i → | z |.(cos Ф + i.sin Ф)
Now let's calculate z:

z² = a²+b² → z² = 0² (-2.i)² → z = -2(i)² →z= -2(-1) → z = |2|
tan Ф = b/a = -2/0 → tan Ф → - ∞ ↔ Ф = -90° or Ф = 270°
then the - 2 I ↔ |2|(cos 270° + i.sin 270°)

7 0

3 years ago

Is -4 rational or irrational

Hoochie [10]

It’s a Rational number

8 0

3 years ago