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KatRina [158]
3 years ago
15

Find the area of the rectangle 2 1/2cm 4cm

Mathematics
1 answer:
Neporo4naja [7]3 years ago
5 0
2.5•2.5= 5 , 4•4=8 , 8+5=12
You might be interested in
A square building with an area of 225 m2 has a garden surrounding it that has an equal width on all sides. The area of the garde
xenn [34]

Answer:

The width of the garden is 1.16 m

Step-by-step explanation:

step 1

<em>Find the area of the garden</em>

To find out the area of the garden multiply by 1/3 the area of the building

225(\frac{1}{3})=75\ m^2

step 2

Find the length side of the square building

The area of a square is

A=b^2

where

b is the length side of the square

we have

A=225\ m^2

so

b^2=225\\b=15\ m

step 3

Find the width of the garden

Let

x ----> the width of the garden

we know that

The area of the building plus the area of the garden is equal to

(15+2x)^2=225+75

solve for x

225+60x+4x^2=225+75\\4x^2+60x-75=0

Solve the quadratic equation by graphing

using a graphing tool

The  solution is x=1.16 m

see the attached figure

therefore

The width of the garden is 1.16 m

Find the exact value

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

4x^2+60x-75=0  

so

a=4\\b=60\\c=-75

substitute in the formula

x=\frac{-60\pm\sqrt{60^{2}-4(4)(-75)}} {2(4)}

x=\frac{-60\pm\sqrt{4,800}} {8}

x=\frac{-60\pm40\sqrt{3}} {8}

x=\frac{-60+40\sqrt{3}} {8}\ m  ----> exact value

8 0
3 years ago
Please help me please
balu736 [363]

Answer:

Step-by-step explanation:

2n + 8.5 = 23

2n = 14.5

2n = 7.25 or 7 1/4 feet

3 0
3 years ago
Find the probability that a point chosen at random will lie in the shaded area.
son4ous [18]
I think the answer is D
6 0
3 years ago
Which of the following relationships is a direct variation?
mafiozo [28]

Answer:Well All i know is that the answer is not C I'm sorry I dont know the correct answer but I know for a fact that the answer isn't C

Step-by-step explanation:

4 0
4 years ago
The sum of 30 times (1/3)^(n-1) from 1 to infinity
sergeinik [125]

Let

S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}

Then

\dfrac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^{n+1}}

and

S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}

and as n\to\infty, we end up with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32

So we have

\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45

8 0
3 years ago
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