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3 years ago

Can someone please help me solve this

Mathematics

1 answer:

Lady_Fox [76]3 years ago

4 0

The "parent function" is y = (log to the base 2 of) x

The domain of this function is (0, infinity) (all real numbers greater than zero).

The range of this function is the same as above.

If you replace "x" with "x+1" in the parent function, the associated graph will look the same as that of the given function, EXCEPT that it will be translated by 1 unit to the left.

After this has happened, that "-3" will shift the entire new graph downward by 3 units.

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2 years ago

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2 years ago

P and q are complex numbers such that ∣p∣ = 2/5 x + 2 , ∣q∣ = −3/4 x − 1 , and ∣p+q∣ = 2x + 4 .

Leya [2.2K]

Answer: (b)

Step-by-step explanation:

Given

$\left | p\right |=\dfrac{2}{5}x+2$

$\left | q\right |=\dfrac{-3}{4}x-1$

$\left | p+q\right |=2x+4$

for two complex number, $z_1,z_2$

Apply the above the property

$\Rightarrow 2x+4\leq \dfrac{2}{5}x+2-\dfrac{3}{4}x-1\\\\\Rightarrow 2x+4\leq \dfrac{8x-15x}{20}+1\\\\\Rightarrow 2x+3\leq -\dfrac{7x}{20}\\\Rightarrow x\leq -\dfrac{60}{47}$

Also, the absolute value of each complex number must be greater than or equal to zero

$\text{Case-1}\\\\\Rightarrow \dfrac{2}{5}x+2\geq 0\\\\\Rightarrow x\geq -5\\\\\text{Case-2}\\\\\Rightarrow -\dfrac{3}{4}x-1\geq 0\\\\\Rightarrow x\leq -\dfrac{4}{3}\\\\\text{Case-3}\\\\\Rightarrow 2x+4\geq 0\\\Rightarrow x\geq -2$

Taking the intersection of the above values of x, we get

$x\in [-2,-\dfrac{4}{3}]$

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2 years ago