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Mariulka [41]
4 years ago
11

The first side of a triangle is 8 m shorter than the second side. The third side is 4 times as long as the first side. The perim

eter is 26 m. Find the length of each side.
Mathematics
2 answers:
anyanavicka [17]4 years ago
4 0
<span>1st side is = x-8
2nd side is = x
3rd side is = 4(x-8)
</span><span>Equations - x-8+x+4(X-8)=28
x-8+x+4x-32=28
6x-40=28
6x=68
x = Does is not an equal number.</span>hope this helps, let me know if its right?
Fynjy0 [20]4 years ago
3 0
All you need to do is half base times high 
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Anastaziya [24]

Answer:

the answer is X

Step-by-step explanation:

it is x

6 0
3 years ago
In the diagram of a quadrilateral below, the variables represent the length of the sides in inches.
SVETLANKA909090 [29]
Add 11 + 16. Plug in 11 into b-2 since b equals eleven. You should now have 11+16+11-2. Do the same thing and plug in 16 for c and 11 for b for c-b. Your solution will now be 11+16+11-2+16-11.
4 0
3 years ago
What is the expansion of (3+x)^4
Vlad1618 [11]

Answer:

\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81

Step-by-step explanation:

Considering the expression

\left(3+x\right)^4

Lets determine the expansion of the expression

\left(3+x\right)^4

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=3,\:\:b=x

=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i

Expanding summation

\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}

i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0

i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1

i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2

i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3

i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

as

\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81

\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x

\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2

\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3

\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4

so equation becomes

=81+108x+54x^2+12x^3+x^4

=x^4+12x^3+54x^2+108x+81

Therefore,

  • \left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81
6 0
3 years ago
Heeeeeeeeeeeeeeeeeeepl
DIA [1.3K]

Answer:

Hi there!

Your answer is;

a)

i) 400% of 240

240 is 100%

× 4

960= 400%

ii) 40% of 240

100% = 240

/100

1% = 2.4

× 40

40% = 96

iii) 4% of 240

100% is 240

/100

1% = 2.4

× 4

4% = 9.6

iv) .04% of 240

100% = 240

/100

1% = 2.4

/100

.01% = .024

× 4

.04% = .096

b) the patterns is that all these numbers equal sometime 96. Each of these have a different decimal place, but have the same actual numbers.

c) 4000% = 240

take the pattern:

400% is 960

Scale it up to 4000 by 10

400% is 960

× 10

4000% is 9600

Hope this helps!

5 0
3 years ago
Type your answer in the box. <br><br> y = 10 b = 6
blondinia [14]

Answer:

The answer is -60 (I believe)

Step-by-step explanation:

3(-10)-5(6)

-30-30 = -30+-30

-30+-30= -60

The answer is -60

6 0
3 years ago
Read 2 more answers
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