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3 years ago

Which is equivalent to 80^1/4 x?

Mathematics

1 answer:

Jlenok [28]3 years ago

4 0

Answer:

B) 4 square root 80^x

Step-by-step explanation:

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Simplify the expression -3÷(-2/5)

Luden [163]

Answer:

7 1/2

Step-by-step explanation:

Because you have to keep, change, flip and it is positive because they both are negative

Can you mark me brainliest

7 0

3 years ago

The radius of a circular puddle is growing at a rate of 25 cm/s.

NikAS [45]

Answer:

a) 2500π cm/s

b) 300√π cm/s

Step-by-step explanation:

Given:

Rate of growth of radius, $\frac{dr}{dt}$ = 25 cm/s

Area of circle is given as:

A = πr²

a)Rate of growth of area, $\frac{dA}{dt}=\frac{d(\pi r^2)}{dt}$

⇒ $\frac{dA}{dt}=(2)\pi r\frac{dr}{dt}$ ............(1)

at r = 50 cm

on substituting the respective values, we get

⇒ $\frac{dA}{dt}=(2)\pi r\frac{dr}{dt}$

⇒ $\frac{dA}{dt}$ = 2π(50)25 = 2500π cm/s

b) when area , A = 36 cm²

36 = πr²

r = $\frac{6}{\sqrt{\pi}}$

thus, using (1)

⇒ $\frac{dA}{dt}=(2)\pi r\frac{dr}{dt}$

on substituting the respective values, we get

⇒ $\frac{dA}{dt}=(2)\pi (\frac{6}{\sqrt{\pi}})25$

⇒ $\frac{dA}{dt}$ = 300√π cm/s

6 0

3 years ago

Please help prove these identities!

lord [1]

<h3>Hi! It will be a pleasure to help you to prove these identities, so let's get started:</h3>

We have the following expression:

$tan(\theta)cot(\theta)-sin^{2}(\theta)=cos^2(\theta)$

We know that:

$cot(\theta)=\frac{1}{cot(\theta)}$

Therefore, by substituting in the original expression:

$tan(\theta)\left(\frac{1}{tan(\theta)}\right)-sin^{2}(\theta)=cos^2(\theta) \\ \\ \\ Simplifying: \\ \\ 1-sin^2(\theta)=cos^2(\theta)$

We know that the basic relationship between the sine and the cosine determined by the Pythagorean identity, so:

$sin^2(\theta)+cos^2(\theta)=1$

By subtracting $sin^2(\theta)$ from both sides, we get:

$\boxed{cos^2(\theta)=1-sin^2(\theta)} \ Proved!$

We have the following expression:

$\frac{cos(\alpha)}{cos(\alpha)-sin(\alpha)}=\frac{1}{1-tan(\alpha)}$

Here, let's multiply each side by $cos(\alpha)-sin(\alpha)$ :

$(cos(\alpha)-sin(\alpha))\left(\frac{cos(\alpha)}{cos(\alpha)-sin(\alpha)}\right)=(cos(\alpha)-sin(\alpha))\left(\frac{1}{1-tan(\alpha)}\right) \\ \\ Then: \\ \\ cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{1-tan(\alpha)}$

We also know that:

$tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}$

Then:

$cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{1-\frac{sin(\alpha)}{cos(\alpha)}} \\ \\ \\ Simplifying: \\ \\ cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)}} \\ \\ Or: \\ \\ cos(\alpha)=\frac{\frac{cos(\alpha)-sin(\alpha)}{1}}{\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)}} \\ \\ Then: \\ \\ cos(\alpha)=cos(\alpha).\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)-sin(\alpha)} \\ \\ \boxed{cos(\alpha)=cos(\alpha)} \ Proved!$

We have the following expression:

$\frac{cos(x+y)}{cosxsiny}=coty-tanx$

From Angle Sum Property, we know that:

$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$

Substituting this in our original expression, we have:

$\frac{cos(x)cos(y)-sin(x)sin(y)}{cosxsiny}=coty-tanx$

But we can also write this as follows:

$\\ \frac{cosxcosy}{cosxsiny}-\frac{sinxsiny}{cosxsiny}=coty-tanx \\ \\ Simplifying: \\ \\ \frac{cosy}{siny}-\frac{sinx}{cosx} =coty-tanx \\ \\ But: \\ \\ \frac{cosy}{siny}=coty \\ \\ \frac{sinx}{cosx}=tanx \\ \\ Hence: \\ \\ \boxed{coty-tanx=coty-tanx} \ Proved!$