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3 years ago

Is 5.7 greater than or less than 5.77

Mathematics

2 answers:

weeeeeb [17]3 years ago

8 0

Actually 5.77 is greater than 5.7 because it is closer to 6 than 5.7 if you make a number line

weqwewe [10]3 years ago

4 0

5.7 is greater than 5.77

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In 3a + b - 12 Name The Coefficients. PLZ<br> A) 3<br> B )3 , 1 , 12<br> C) 3, 1<br> D) a , b

lidiya [134]

C. 3, 1 is the answer!

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3 years ago

Read 2 more answers

What is 9 to the power of -3

larisa86 [58]

Answer: 1/729

.........................

3 0

3 years ago

Please please help me

Gekata [30.6K]

Answer:

$\large\boxed{\dfrac{\boxed{8}}{81}}$

Step-by-step explanation:

$\text{If}\ a_1,\ a_2,\ a_3,\ a_4,\ ...,\ a_n\ \text{is the geometric sequence, then}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...=\dfrac{a_n}{a_{n-1}}=constant=r-\text{common ration}.\\\\\text{We have}\ \dfrac{1}{2},\ \dfrac{1}{3},\ \dfrac{2}{9},\ \dfrac{4}{27},\ ...\\\\\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{1}{3}\cdot\dfrac{2}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{9}\cdot\dfrac{3}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{4}{27}}{\frac{2}{9}}=\dfrac{4}{27}\cdot\dfrac{9}{2}=\dfrac{2}{3}$

$\bold{CORRECT}\\\\\dfrac{x}{\frac{4}{27}}=\dfrac{2}{3}\qquad\text{multiply both sides by}\ \dfrac{4}{27}\\\\x=\dfrac{2}{3}\cdot\dfrac{4}{27}\\\\x=\dfrac{8}{81}$

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

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