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mixer [17]
3 years ago
8

how can you tell before you divide 425 by 9 that the first digit of the quotient is in the tens place?

Mathematics
1 answer:
Goshia [24]3 years ago
4 0
I don't understand that question
You might be interested in
Solve by substitution y=x+4, y=2x+5
Amiraneli [1.4K]
X+4=2x+5
-1=x
hope this helps
4 0
2 years ago
Sebastian and his family picked 4 1/4 bushels of red apples and 2 5/6 of green apples at the orchard. How many more red apples w
11111nata11111 [884]

Answer:

1 5/12

Step-by-step explanation:

4 1/4- 2 5/6= 1 5/12

Also   1 5/12+  2 5/6= 4 1/4

8 0
3 years ago
Find the center and radius of the circle whose equation is x 2 + y 2 + 10x − 1
Margaret [11]

we conclude that the center of the circle is the point (-5, 0).

<h3>How to find the center of the circle equation?</h3>

The equation of a circle with a center (a, b) and a radius R is given by:

(x - a)^2 + (y - b)^2 = R^2

Here we are given the equation:

x^2 + y^2 + 10x - 1 = 0

Completing squares, we get:

x^2 + 10x + y^2 = 1

Now we can add and subtract 25 to get:

x^2 + 10x + 25 - 25 + y^2 = 1\\\\(x + 5)^2 -25 + y^2 = 1\\\\(x + 5)^2 + (y - 0)^2 = 26

Comparing that with the general circle equation, we conclude that the center of the circle is the point (-5, 0).

If you want to learn more about circles:

brainly.com/question/1559324

#SPJ1

3 0
2 years ago
What is the cost of 20pens is rs 1200,what is the cost of one pen​
myrzilka [38]

Answer:

rs 60

1200/20=60......

3 0
2 years ago
HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH
8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

4+\sqrt{6x}

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}

Thank you

8 0
3 years ago
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