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3 years ago

What is the place value for 3 in 6,035

Mathematics

1 answer:

bonufazy [111]3 years ago

5 0

The place value in 6,035 is 30

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Rewrite the following differences of squares as a product of two integers. (a). 81 − 1 (b). 400 − 121

Hatshy [7]

Answer:

(a) 10×8

(b) 31×9

Step-by-step explanation:

We have given that

(a) 81 - 1

We have to write this in form of product of two integers

We know the algebraic equation $a^2-b^2=(a+b)(a-b)$

As 81 -1 can be written as $9^2-1^2$

So $9^2-1^2=(9+1)(9-1)=10\times 8$

(b) We have given 400 - 121

We have to write this in form of product of two integers

We know the algebraic equation $a^2-b^2=(a+b)(a-b)$

As 81 -1 can be written as ${20}^2-{11}^2$

So ${20}^2-{11}^1=(20+11)(20-11)=31\times 9$

4 0

3 years ago

I'm bo.red here's a question, How to find area of a circle just put the formula

Nuetrik [128]

Answer:

Step-by-step explanation:

A = $\pi$ $x^{2}$

7 0

2 years ago

Read 2 more answers

Is the point 3/5 and 4/5 corresponds to an angle in the unit circle what is the csc

Mariulka [41]

Answer:

csc = 5/4

Step-by-step explanation:

Assuming that the two given points 3/5 and 4/5 are the two sides of a triangle and a right angle 0 with legs x and y and z as its hypotenuse.

Then using Pythagoras Theorem:

$z^2=x^2+y^2$

$z^2=(\frac{3}{5} )^2+(\frac{4}{5} )^2$

$z^2 = \frac{9}{25} + \frac{16}{25}$

$z^2=\frac{25}{25}$

$z^2=1$

Taking square root on both the sides:

$z=1$

Now, $sin\alpha=\frac{4}{5}$

and we know that $csc = \frac{1}{sin\alpha }$

Therefore, csc = 5/4.

8 0

4 years ago

Rewrite without absolute value y=|x−5|+|x+5|, if x<−5 thank you

ioda

y = | x -5| + |x+5|

as range of x is

-5 < x < 5

we can see if we put x = 0 we get

y = 10

we put any value of x in given range first term is always negative without absolute. and 2nd term is always positive

y = | x -5| + |x+5|

= -( x-5) + x + 5

= - x + 5 + x + 5

= 10

y = 10 where -5 < x < 5

8 0

4 years ago

How to solve piecewise functions step by step?

Zanzabum

If you can graph a function, you can graph piecewise functions. Each one of them is a different line or curve within the domain for that specific line of curve. If the domain states it's less than or greater than a number, you circle that point on the line. If the domain states it's also possibly equal to the point at the beginning or the end, you make a closed dot.

I"m sure that's clear as mud, please respond with any additional questions.

5 0

4 years ago