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3 years ago

Ivan, a botanist, treated a 2 cm plant with a special growth fertilizer. With this fertilizer,

Mathematics

1 answer:

chubhunter [2.5K]3 years ago

6 0

Answer

poop

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The function y = x2 is transformed to y = (x + 4)2. Which statement is true about the transformed function? It is an even functi

xxMikexx [17]

Even function: f(-x) = f(x). If you replace x by -x you should find the same function.
Odd function: f(-x) = -f(x). If you replace x by -x you find the same function with opposite sign;
Is f(-x) = f(x)?
f(x) = (x+4)² = x² + 8x +16
f(-x) = (-x+4)² = x² - 8x + 16, then it's not an even function

Is f(-x) = -f(x)?
f(-x) = (-x+4)² = x² + 8x + 16 , then it's not an odd function
It is neither an even nor an odd function

7 0

3 years ago

Pre-Calculus Help? Polar and Parametric Equations?

Paraphin [41]

$\bf x=rcos(\theta )\implies \cfrac{x}{cos(\theta )}=r\\\\
y=rsin(\theta )\implies \cfrac{y}{sin(\theta )}=r
\\\\
-------------------------------\\\\
r=3cos(\theta )\implies \cfrac{x}{cos(\theta )}=3cos(\theta )\implies x=3cos(\theta )cos(\theta )
\\\\\\
\boxed{x=3cos^2(\theta )}
\\\\\\
r=3cos(\theta )\implies \cfrac{y}{sin(\theta )}=3cos(\theta )\implies \boxed{y=3cos(\theta )sin(\theta )}$

7 0

2 years ago

Is anybody else here to help me ??

Akimi4 [234]

Answer:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

$\cot(x)+\tan(x)$

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

Step-by-step explanation:

I'm going to use $x$ instead of $\theta$ because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

$\cot(x)+\cot(\frac{\pi}{2}-x)$

I'm going to use a cofunction identity for the 2nd term.

This is the identity: $\tan(x)=\cot(\frac{\pi}{2}-x)$ I'm going to use there.

$\cot(x)+\tan(x)$

I'm going to rewrite this in terms of $\sin(x)$ and $\cos(x)$ because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities: $\frac{\cos(x)}{\sin(x)}=\cot(x)$ and $\frac{\sin(x)}{\cos(x)}=\tan(x)$

So we have:

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

I'm going to factor out $\frac{1}{\sin(x)}$ because if I do that I will have the $\csc(x)$ factor I see on the right by the reciprocal identity:

$\csc(x)=\frac{1}{\sin(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show $\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}$ is equal to $\csc(\frac{\pi}{2}-x)$ .

So since I want one term I'm going to write as a single fraction first:

$\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}$

Find a common denominator which is $\cos(x)$ :

$\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}$

$\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}$

$\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}$

By the Pythagorean Identity $\cos^2(x)+\sin^2(x)=1$ I can rewrite the top as 1:

$\frac{1}{\cos(x)}$

By the quotient identity $\sec(x)=\frac{1}{\cos(x)}$ , I can rewrite this as:

$\sec(x)$

By the cofunction identity $\sec(x)=\csc(x)=(\frac{\pi}{2}-x)$ , we have the second factor of the right hand side:

$\csc(\frac{\pi}{2}-x)$

Let's just do it all together without all the words now:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

$\cot(x)+\tan(x)$

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

7 0

3 years ago

Help, I still cannot wrap my head around this

atroni [7]

Answer:

9 inches.

Step-by-step explanation:

Fun question! The length of a space diagonal for a cube is $\sqrt{3*sidelength^2}$

Thus, the first cube has a side diagonal of $\sqrt{3}$

The next cube would have a side diagonal of 3

The third cube would have a side diagonal of $3\sqrt{3}$

And the fourth cube would have a side diagonal of 9.

Since the fifth cube has a side length equal to the side diagonal of the fourth cube, it should be 9 inches.

7 0

3 years ago

What is the sale price of a car that has an original price of 15,500 with a 14% off sale

Mars2501 [29]

14% of $15,500 is $2,170. $15,500 minus $2,170 is $13,330.

When the car is on sale, the price is reduced from $15,500 to $13,330.

5 0

3 years ago