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aleksklad [387]
3 years ago
9

Factorise 5x*2-23*x+12

Mathematics
1 answer:
egoroff_w [7]3 years ago
3 0

5x²-23x+12

5x²-20x-3x +12

5x(x-4)-3(x-4)

(x-4)(5x-3)

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Question is show below.
WITCHER [35]

(1/4ab)^-2

Remove negative exponent by rewritten the equation as (4ab)^2

AUse Product rule to get:

4^2a^2b^2

Now raise 4 to the second power to get:

16a^2b^2


The last answer is correct.

7 0
3 years ago
Can someone help with this step by step ty!
Alik [6]

Answer:

Area = 88 units squared

Perimeter = 40 units

Step-by-step explanation:

The formula for the Area of the parallelogram is base (11) times height (8)

The perimeter is just adding all the outsides together (11 + 11 + 9 + 9)

7 0
3 years ago
What is -5/6 time -6/5
AURORKA [14]

Answer:

1

Step-by-step explanation:

you are just multiplying by the reciprocal here which would make this equal to 1

4 0
3 years ago
In a certain urban are, the relationship between , number of students ,x, in thousands, and the number schools, y, has an x-inte
chubhunter [2.5K]

Answer:

-12 = 3x - 4y

Step-by-step explanation:

Two points on the graph are (-4, 0) and (0, 3).  Moving from the first point to the second, we see x (the 'run') increase by 4 and y (the 'rise') increase by 3.  Thus, the slope of this line is m = rise / run = 3/4.

Using the point slope form, we get:

y - 3 = (3/4)(x - 0), or

4y - 12 = 3x, or

-12 = 3x - 4y (which is in Standard Form).

6 0
3 years ago
Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.
skad [1K]

Answer:

The absolute maximum and minimum is 20\; \text{and} -20.

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

f_x=y-4=0

f_y=x=0

The only solution to this system of equations is the point (0, 4), which lies in the domain.

f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1

\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1

\therefore (0,4) is a saddle point.

Boundary points -  (4,0),  (-4,0), (0,16)

Along boundary  y=16-x^2

   f=x(16-x^2)-4x

=16x-x^3-4x

\Rightarrow f^'=16-3x^2-4=0

\Rightarrow 3x^2=12

\Rightarrow x=\pm2,\;\;y=14

Values of f(x) at these points.

\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}

Therefore, the absolute maximum and minimum is 20\; \text{and} -20.

6 0
4 years ago
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