Value of m{8+[b(m-1)+3]} when m = -2 and b = 2/3

Mathematics

1 answer:

Mrrafil [7]3 years ago

3 0

Answer:

-18

Step-by-step explanation:

m = -2 Start with the expression that is within the most number of brackets.

m - 1 = -2 - 1 = -3

Multiply that result by 2/3

-3 * 2/3 = - 2

To that result add 3 which will get you through the square brackets.

- 2 + 3

+ 1

Just outside the square bracket is 8 Eight should be added to 1

8 + 1 = 9

Now you multiply that result by - 2

-2 * 9

= - 18

The answer is 6

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4/3(6y+21)-17>43 help pls

dezoksy [38]

Answer:

Step-by-step explanation:

4/3(6y+21)-17=43

move all terms to the left:

4/3(6y+21)-17-(43)=0

Domain of the equation: 3(6y+21)!=0

y∈R

We add all the numbers together, and all the variables

4/3(6y+21)-60=0

You multiply all the terms by the denominator

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multiply elements

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3 years ago

What is the best estimate for 8.3 x 4.9?

sammy [17]

Answer:

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Step-by-step explanation:

I calculated it

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3 years ago

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.