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Anika [276]
3 years ago
6

(4ab - 8a 2 + 12ac) ÷ 4a

Mathematics
1 answer:
ioda3 years ago
7 0

we are given

\frac{(4ab-8a^2+12ac)}{4a}

we can see that

4a is in multiple in numerator

so, we can factor it out

\frac{4a(b-2a+3c)}{4a}

now, we can cancel 4a

and we get

(b-2a+3c).............Answer

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Answer:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
Help with some math homework? The area of a triangle is 110 square inches. the height of the triangle is 8 less than 3 times its
Alenkasestr [34]
A = h x b; where A = area of a triangle; h = height of the triangle; b = base of the riangle;
So, we have the equation : 110 = (3b - 8)b;
We solve the equation 3b^2 - 8b - 110 = 0;
The positiv solution  is b = 7.53 cm;
h = 3 x 7.53 - 8;
h = 14.59 cm;
3 0
3 years ago
I NEED HELP ON THIS PLZZZ HELPP
adell [148]

Answer:

76 \:  {in}^{2}

Step-by-step explanation:

Area of the figure = Are of square with side 8 in + 2 times the area of one triangle with base (8 - 5 = 3) 3 in and height 4 in

=  {(8)}^{2}  + 2 \times  \frac{1}{2}  \times 3 \times 4 \\  \\  = 64 + 12 \\  \\  = 76 \:  {in}^{2}

3 0
3 years ago
Write parametric equations of the line through the points (7,1,-5) and (3,4,-2). please use the first point as your base-point w
Roman55 [17]

Given:

A line through the points (7,1,-5) and (3,4,-2).

To find:

The parametric equations of the line.

Solution:

Direction vector for the points (7,1,-5) and (3,4,-2) is

\vec {v}=\left

\vec {v}=\left

\vec {v}=\left

Now, the perimetric equations for initial point (x_0,y_0,z_0) with direction vector \vec{v}=\left, are

x=x_0+at

y=y_0+bt

z=z_0+ct

The initial point is (7,1,-5) and direction vector is \vec {v}=\left. So the perimetric equations are

x=(7)+(-4)t

x=7-4t

Similarly,

y=1+3t

z=-5+3t

Therefore, the required perimetric equations are x=7-4t, y=1+3t and z=-5+3t.

5 0
3 years ago
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