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4 years ago

Increase £2 by fifty percent.(Show working out)

2 answers:

7 0

If I were to increase 2 pounds by fifty percent, I would increase it by half of its value.
Half of 2 is 1, therefore I would increase 2 pounds by 1, thus increasing it to a grand total of 3 pounds. I hope this helps, have a nice day, and if you need more help just ask! :)

uranmaximum [27]4 years ago

6 0

First, find out the 50% of £2 then add it to £2 for final value.

It would go in like this -

Increment =50% of £2
= 50/100 × 2
= £1

Now, final value = £1 + £2 = £3

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There are 7 more girls than boys in a class. There are 23 students in the class in all how many girls are there in the class. Wh

Flura [38]

The unknown is how many girls and boys are in the class.

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3 years ago

the function M(x) -1/45x^2+2x-20 where 30 65 represents the miles per gallon used by a certain brand of car at a speed of x mile

JulsSmile [24]

The answer would be option B

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3 years ago

8-35. Use your knowledge of polygons to answer the questions below, if possible.

Nataliya [291]

Answer:

(a)I. 13 Sides II. 7 Sides

(b) 4 Sides, Square

(c)x=52 degrees

(d)x=107 degrees

(e)1620 degrees

Step-by-step explanation:

(a)The Sum of the Interior angle of polygon with n sides is derived using the formula: (n-2)180.

I. If the Interior angle is $1980^0$

Then:

$(n-2)180^0=1980^0\\$Divide both sides by 180^0 $ to isolate n$\\n-2=11\\$Add 2 to both sides of the equation\\n-2+2=11+2\\n=13$

The polygon has 13 sides.

II. If the Interior angle is $900^0$

Then:

$(n-2)180^0=900^0\\$Divide both sides by 180^0 $ to isolate n$\\n-2=5\\$Add 2 to both sides of the equation\\n-2+2=5+2\\n=7$

The polygon has 7 sides.

(b)The sum of the exterior angle of a polygon is 360 degrees,

Each exterior angle of a n-sided regular polygon is: $\frac{360^0}{n}$

If the exterior angle of a regular polygon is 90°

Then:

$90\°=\frac{360^0}{n}\\ 90n=360\\n=4$

The regular polygon has 4 sides and it is called a Square.

(c)The Sum of the Interior angle of polygon with n sides is derived using the formula: (n-2)180.

Each Interior angle of a regular n-sided polygon is: $\frac{(n-2)180^0}{n}$

For a pentagon, n=5

Then:

$\frac{(5-2)180^0}{5}=2x+4\\108=2x+4\\108-4=2x\\104=2x\\x=52^0$

(d)The sum of the exterior angle of a polygon is 360 degrees.

If four of the exterior angles of a pentagon are 57, 74, 56, and 66.

Let the fifth angle=x

Then:

$57+74+56+66+x=360^0\\253+x=360\\x=107^0$

(e)The Sum of the Interior angle of polygon with n sides is derived using the formula: (n-2)180.

In an 11-gon., n=11

Therefore, the sum of the interior angle=(11-2)180= $1620^0$

The sum of the interior angle does not change either in a regular or irregular polygon.

3 0

3 years ago

What is the square root of 16?

natulia [17]

square root of 16 would be 4

4 x 4 = 16

8 0

4 years ago

Read 2 more answers

2x10^-4 - 1x10^-5 show work. I got an answer of 1.9x10^-4,

oksian1 [2.3K]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------$

$\bf 2\times 10^{-4}-1\times 10^{-5}\implies 2\cdot \cfrac{1}{10^4}-1\cdot \cfrac{1}{10^5}\impliedby 
\begin{array}{llll}
\textit{using the LCD}\\
of~10^5
\end{array}
\\\\\\
\cfrac{2}{10^4}-\cfrac{1}{10^5}\implies \cfrac{(10\cdot 2)~-~(1\cdot 1)}{10^5}\implies \cfrac{20-1}{10^5}\implies \cfrac{19}{10^5}
\\\\\\
19\cdot \cfrac{1}{10^5}\implies 19\times 10^{-5}$

8 0

3 years ago

Read 2 more answers