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3 years ago

14

Why is this function nonlinear?

1 answer:

Neporo4naja [7]3 years ago

3 0

Answer:

Function is nonlinear because Linear functions have a constant slope, so nonlinear functions have a slope that varies between points

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Plz answer last question and im lost!

harkovskaia [24]

Answer:

$\pi$ radian

Step-by-step explanation:

We know that angle for a full circle is 2 $\pi$

In the given figure shape is semicircle

hence,

angle for semicircle will be half of angle of full circle

thus, angle for given figure = half of angle for a full circle = 1/2 * 2 $\pi$ = $\pi$

Thus, answer is $\pi$ radian

alternatively, we also know that angle for a straight line is 180 degrees

and 180 degrees is same as $\pi$ radian.

6 0

3 years ago

PLEASE NO SPAM!!!! I REALLY NEED TO GET A GOOD GRADE ON THIS!

kenny6666 [7]

Answer:

I think it is A

Step-by-step explanation:

3 0

2 years ago

A pharmacist who wants to mix 836% solution with a 12% saline solution to get 64 mL at 33% saline solution how much of each solu

GenaCL600 [577]

<span>x = 60 ml of the 10% solution </span>

7 0

3 years ago

The radius of a semicircle is 3 kilometres. What is the semicircle's perimeter?

alina1380 [7]

Answer:

In terms of Pi: $3\pi + 6$ km

Exact value (rounded): 15.42477 km

Approximated (3.14): 15.42 km

Step-by-step explanation:

Hello!

A semicircle's perimeter can be found using the formula: $P = \pi r + 2r$

We can plug in the values to solve for the perimeter.

<h3>Solve</h3>

$P = \pi r + 2r$
$P = \pi(3)+ 2(3)$
$P = 3\pi + 6$

In terms of Pi, it is $3\pi + 6$ ,

Exact value of Pi, it is 15.42477

Approximation of Pi (3.14), it is = 15.42

6 0

2 years ago

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

3 years ago