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3 years ago

Hello chaps, please help me with this absurb eqution, "I think of a number add one then double the result

Mathematics

1 answer:

ziro4ka [17]3 years ago

6 0

Answer:

2n+2

Step-by-step explanation:

Let the number = n.

Add 1 to n:

n+1

Double n+1:

2(n+1)

Distribute:

2n+2

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How many solutions does the equation have? 7x +3 - x=3(1 + 2x) (A) 0 (B) 1. (C) infinitely many

tekilochka [14]

Answer:

C) Infinitely many

Step-by-step explanation:

7x+3-x=3(1+2x)

6x+3=3+6x

Subtract 3 from both sides

6x = 6x

Infinitely many solutions

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3 years ago

Aaden is packing boxes into a carton that is 8 inches long 8 inches wide and 4 inches tall. the boxes are 2 inches long 1 inch w

vredina [299]

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3 years ago

Please help!! MathsWatch, Surd Expressions, Clip 207c

dexar [7]

Answer:

$A=(72+36\sqrt{3})\ mm^2$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

Let

a ----> the height of rectangle in mm

b ---> the base of rectangle in mm

step 1

Find the base of rectangle

$b=AB+BC+CD+DE$ ----> by segment addition postulate

substitute

$b=3+3+3+3=12\ mm$

step 2

Find the height of rectangle

$a=FB+CG+GH$ ---> by segment addition postulate

substitute the given values

$a=3+CG+3\\a=6+CG$

Find the length sides CG

Applying the Pythagorean Theorem

$BG^2=BC^2+CG^2$

substitute the given values

$6^2=3^2+CG^2$

$CG^2=6^2-3^2$

$CG=\sqrt{27}\ mm$

simplify

$CG=3\sqrt{3}\ mm$

therefore

$a=(6+3\sqrt{3})\ mm$

step 3

Find the area of rectangle

$A=ab$

we have

$a=(6+3\sqrt{3})\ mm$

$b=12\ mm$

substitute

$A=(6+3\sqrt{3})(12)=(72+36\sqrt{3})\ mm^2$

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3 years ago

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jeka57 [31]

Answer:

6/14 or in its simplest form 3/7

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3 years ago

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Sati [7]

I think the answer is 37.7

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3 years ago