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3 years ago

What is 52cm as a fraction of 90cm? (as a simplified fraction)

Mathematics

1 answer:

katrin2010 [14]3 years ago

5 0

Answer:

$\frac{26}{45}$ and $\frac{3}{11}$

Step-by-step explanation:

(a)

Expressing as a fraction

$\frac{52}{90}$ ( divide numerator/ denominator by 2 )

= $\frac{26}{45}$ ← in simplest form

(b)

Expressing as a fraction

$\frac{30}{110}$ ( divide numerator/ denominator by 10 )

= $\frac{3}{11}$ ← in simplest form

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Determine the domain and range of the function f(x)= 3x+2. Also, state the intervals where the function f(x)= 3x+ 2 is increasin

aev [14]

Answer:

Increasing on it's domain $(-\infty,\infty)$ because the slope is positive.

The domain and range are both all real numbers, also known as

$(-\infty,\infty)$ .

Step-by-step explanation:

All domain really means is what numbers can you plug in and you get number back from your function.

I should be able to plug in any number into 3x+2 and result in a number. There are no restrictions for x on 3x+2.

The domain is all real numbers.

In interval notation that is $(-\infty,\infty)$ .

Now the range is the set of numbers that get hit by y=3x+2.

Well y=3x+2 is a linear function that is increasing. I know it is increasing because the slope is positive 3. I wrote out the positive part because that is the item you focus on in a linear equation to determine if is increasing or decreasing.

If slope is positive, then the line is increasing.

If slope is negative, then the line is decreasing.

So y=3x+2 hits all values of y because it is increasing forever. The range is all real numbers. In interval notation that is $(-\infty,\infty)$ .

3 0

3 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$