What is 21,253 to the nearest hundreds

Mathematics

1 answer:

RUDIKE [14]3 years ago

4 0

21,253 it stays the same cause it’s not over 5

You might be interested in

Write down the integer values that satisfy the inequality -5<_x<3

Naddik [55]

Answer:

(-4,-3,-2,-1,0,1,2)

Step-by-step explanation:

..........i hope it is............

8 0

3 years ago

Can y’all help me on number 20

Gwar [14]

It’s a brainliest me pls

6 0

3 years ago

Read 2 more answers

Write a expression that shows 3 times the sixth power of 10

user100 [1]

(10 raised to the power of 6)×3

(10*6) ×3

8 0

3 years ago

Write an equation for the translation of <img src="https://tex.z-dn.net/?f=y%5Cfrac%7B3%7D%7Bx%7D" id="TexFormula1" title="y\fra

lisov135 [29]

<h2><u>Answer:</u></h2><h2> $\boxed{\boxed{y=\frac{3}{x-6}-6}}$ </h2><h2 /><h2><u>Solution Steps:</u></h2>

______________________________

<h3>1.) Substitute x, y, and general form using the formulas:</h3>

Formula for x: $x=d=6$

Formula for y: $y=k=-6$

Formula for general form y: $y=\frac{a}{x-d}+k=y=\frac{3}{x-6}+(-6)$

Note:

<em> - When doing problems like this, it's best to list out each formula and do the drag down method, and replace methods rather than doing it the entire mathematical way, as doing it that way takes a long time and it's more confusing than it should be.</em>

<u>Equation at the end of Step 1:</u>

$y=\frac{3}{x-6}+(-6)$

<h3>2.) Simplify the d and k forms:</h3>

$y=\frac{3}{x-6}+(-6)=y=\frac{3}{x-6}-6$

<em> - The x(d) stays the same since the 6 was positive and we plugged it in making it negative, so it doesn't need to be changed. (Another way to know if you need to change it is if the original </em> $(d)$ <em> or </em> $(k)$ <em> when plugged in, you had to put them in parenthesis then most likely it needs to be changed. </em>

<em> - The y(k) changes from </em> $+(-6)$ <em> to </em> $-6$ <em> because when a </em> $(+)$ <em> and </em> $(-)$ <em> are next to each other, the rule is </em> $+$ <em>. The Minus always wins in this type of problem situation.</em>