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3 years ago

11

Square root of 244 by division method

1 answer:

navik [9.2K]3 years ago

7 0

Answer:

The square root of 244 is 15.62 ( up-to 2 decimal places)

Step-by-step explanation:

Following is the division method for square root of 244 with rules as follows:

Make pairs of the given digit starting from right (unit place). They will be calculated one by one as formed in pairs/singles.
Digit added for divisor and quotient will be same at each step.
Each next dividend will be formed by subtracting product of divisor and quotient from first digit/pair.
Each next divisor will be formed by taking two times of previous quotient and annexing it with suitable digit that gives equal to or less than the dividend.

Following solution gives square root of 244:

I hope it will help you!

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16pts for help with this math question

diamong [38]

Y=76 x-18
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3 years ago

Read 2 more answers

Guys i need ur help fast please

Maslowich

Answer:

a) -3

b) 5

c) 7

d) 4

Step-by-step explanation:

We have the function $p(x)=5x^3-3x^2+2x^5+7$

a) We need to find the coefficient of $x^2$ .

This means that we need to find out what number is alongside the $x^2$ in this equation. From the function, we can find that it is -3

b) Now we need to find the degree of $p(x)$ . Recall that the degree refers to the highest power of x that is present. As $x^5$ is the largest power, our degree would be 5.

c) The constant term refers to the number within the function that does not have any x's with it. In this function, that number would be 7.

d) Now we need to find the number of terms. For this one, we just need to count how many terms are separated by + or - signs. There are 4 in this function.

6 0

3 years ago

Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

4 0

3 years ago

Here try to figure out this problem

Marysya12 [62]

Answer:

2

Step-by-step explanation:

2 is the slope, it's going up 2, and right 1 which is 2/1 or just 2

Hope this helps

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3 years ago

What is the value of X?<br><br> x=

Sever21 [200]

Answer:

$x=12$

Step-by-step explanation:

All of the sides are the same length.

$5x-22=4x-10=3x+2$

$5x-22=4x-10$

$x=12$

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3 years ago