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saw5 [17]
3 years ago
13

please help i am doing a test. Explain how you can use partial quotients to divide 7,563 ÷ 4 = ________.

Mathematics
1 answer:
Elza [17]3 years ago
6 0

Answer:

1890.75

so this is the answer to your question  

You might be interested in
An expression is shown below:
PolarNik [594]
Notation
I imagine that the expression you are asked to work with is:
3 x^{3}y+15xy-9 x^{2} y-45y

When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.

PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.

This gives us: 3 x^{3}y+15xy-9 x^{2} y-45y=3y( x^{3}+5x-3 x^{2} -15)

To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.

PART B
Start with the answer in part A. Namely, 3y( x^{3}+5x-3 x^{2} -15). For now let's focus only on what is in the parenthesis. We have four terms so let's take them two at a time. I am separating the expression in two using square brackets. [( x^{3}+5x)]-[3 x^{2} -15]

Let's next factor what is in each bracket:
[( x^{3}+5x)]-[3 x^{2} -15] = [x( x^{2} +5)]-[3( x^{2} +5)]

Notice that both brackets have the same expression in them so now we factor that out: [x( x^{2} +5)]-[3( x^{2} +5)] = (x-3)( x^{2} +5)

Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get: 3 x^{3}y+15xy-9 x^{2} y-45y =(3y) (x-3)( x^{2} +5)

8 0
3 years ago
Please help with this. Thank you very much.
Harman [31]
Here's a graph with 4 points shown.

4 0
3 years ago
I need the solution
shutvik [7]

\dfrac{8c^3}{6(c+d)}\div\dfrac{2c^2}{3c+3d}=\dfrac{4c^3}{3(c+d)}\div\dfrac{2c^2}{3(c+d)}\\\\=\dfrac{4c^3}{3(c+d)}\cdot\dfrac{3(c+d)}{2c^2}=\dfrac{4c^3}{2c^2}=2c

8 0
3 years ago
Write an equation parallel to x - 3y = 9 that passes through the point ( 3, -1 )
ivolga24 [154]

Answer:

Step-by-step explanation:

Use the point-slope formula.

y - y_1 = m(x - x_1)   ....   x_1 = 3 and y_1 = - 1

m is the slope to  x - 3y = 9 because (lines are  parallel )

calculate : m   by equation   x - 3y = 9 :

x = 3y +9

divid by : 3      1/3 x = y + 3

y =  (1/3) x - 3      so : m = 1/3

an equation parallel to x - 3y = 9 that passes through the point ( 3, -1 ) is :

y +1 = (1/3)(x - 3)

4 0
3 years ago
11(x-2)=6(x-3) so helppppppppppp
aleksley [76]
0.8 or 4/5 it’s the same thing
7 0
3 years ago
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