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weeeeeb [17]
1 year ago
10

Simplify the expression 13. 8+(-11. 5). Use pencil and paper. How are 13. 8 + (-11. 5) and (-13. 8) +11. 5

Mathematics
1 answer:
VashaNatasha [74]1 year ago
7 0

On simplifying the expression 13.8+ (-11.5) , we get 2.3 . And additions

13.8 + (-11.5) and (-13.8) +  11.5 are numbers will same value and opposite sign.

<h3>What are rules for adding a positive and a negative number?</h3>

For adding a negative and a positive number, use the sign of the larger number and subtract.

For example: (–8) + 2 = -6

Given that we have to solve 13.8 + (-11.5) = 13.8-11.5 (we subtract smaller no from larger and will use the sign of lager number that is + here)

13.8+(-11.5) = 13.8-11.5 =2.3

13.8+(-11.5) simply means that 13.8 -11.5 which is subtraction of 11.5 from 13.8

but (-13.8) +  11.5 = -13.8 +11.5  =  -( 13.8 -11.5) [taking -1  common from the expression]

Therefore , -13.8 + 11.5 has same value as 13.8- 11.5 but opposite  in sign.

Learn, More about addition of numbers from here:'

brainly.com/question/14092461

#SPJ1

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Number of solutions to a system of equations algebraic. How many solutions does the system have? Can someone help me please....​
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Step-by-step explanation:

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<h2><u>Diagram</u><u>:</u><u>-</u></h2>

\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A}\put(6,0.5){\sf B}\put(1.4,4.3){\sf D}\put(6.6,4.3){\sf C}\end{picture}

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

The diagonals of the Rhombus= d1 & d2=6cm and 8cm

Let the Side=a

As we know that in a Rhombus

{\boxed{\sf side=\sqrt {\left ({\dfrac{d1}{2}}\right)^2+\left ({\dfrac {d2}{2}}\right)^2}}}

  • Substitute the values

{:}\longmapsto\sf a=\sqrt {\left ({\dfrac {6}{2}}\right)^2+\left ({\dfrac {8}{2}}\right)^2 }

{:}\longmapsto\sf a=\sqrt {(3)^2+(4)^2}

{:}\longmapsto\sf a=\sqrt {9+16}

{:}\longmapsto\sf a=\sqrt{25}

{:}\longmapsto\sf a=5cm

<h2>_____________</h2><h3>Again </h3>

we know that in a Rhombus

\boxed{\sf Perimeter=4a}

  • Substitute the values

{:}\longmapsto\sf Perimeter=4×5

{:}\longmapsto\sf Perimeter=20cm

\thereforePerimeter of the rectangle is 20cm.

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